Sheldon  <£•  Company's  Text-l^ooks 


FRENCH   AND   GERMAN 


PEOF.  EEETELS'  NEW  FEENCH  SBEIES. 

The  Oral  Method  with  tite  French,  By  Prof.  Juax 
GcSTAVE  KEEn"ELS,  Author  of  "  Keetels'  New  Method  with  the 
French."     lu  three  parts,  ICtao,  cloth, 

[Tlie  studer.i  is  sctccd  Vic  expense  cf  a  large  hook  in  commencing 
the  study] 

The  Oral  :\Icthod  of  Teaching  livis-  lang^iages  U  fnpcricr  to  ell  others  In 
ir.anv  respects. 

li  teacht!*  the  pupil  to  speak  the  language  he  is  Icamin;;,  end  he  begins  to  do 
to  from  The  first  le<son. 

He  never  becomes  tired  of  the  book,  becaupo  he  feels  that,  with  moderate 
eJofts.  he  is  making  constant  and  rapid  progress. 

The  lessons  are  arranged  so  as  to  bnng  in  one  difficulty  at  a  time.  They  are 
adapted  to  class  purposes,  and  suitable  for  large  or  small  classes,  and  for 
scholars  of  all  ages. 

The  teacher,  with  thi?;  book  in  his  hand,  is  never  at  a  loss  to  profitably 
entertain  his  pupils,  without  rendering  their  task  irksome. 

lu  fine,  the  Oral  Method  works  charmingly  in  a  class.  Teachers  and  pupils 
arc  equally  pleased  with  it :  the  latter  all  learn— the  quick  and  the  dull-  each  in 
proportion  to  his  ability  and  application.  It  is  onr  opinion  that  before  long 
thi.  Oral  Method  will  linci  its  way  into  every  school  where  French  is  taught. 


I  find  that  pnpih  understand  pnd  improve  more  rapidly  tinder  the  Oral 
t hod  of  Keetels' instruction  tha  ^'       ^       -  '  ^ --     •    •" 

E.mvxKKl  Seminary,  Glenn's  Falls, 


?  let  hod  of  Keetels'  instruction  than  any  other  heretofore  used."— A.  Tatix», 

A',  r. 


A  Xcw  Method  of  Learning  the  French  Language. 

By  Jean  Gustate  Keetiils,  Professor  of  French  and  German 
in  the  Brooklyn  Polytechnic  Institute.     12mo. 
A  Keij  to  the  above.    By  J.  G.  Keetet.s. 

This  work  contains  a  clear  nnd  methodical  expose  of  the  principles  of  the 
language,  on  a  plan  entirely  new.  The  arrangement  is  admirable.  The  les- 
sons ere  of  a  suitable  length,  and  within  the\omprehension  of  all  classes  of 
students.  The  exercises  are  various,  and  well  adapted  to  the  purpose  for 
which  they  are  intended,  of  readincr.  writing,  and  speaking  the  l.incru.'iEre.  The 
Grammar  part  is  complete,  and  accompanied  by  questions  and  exi-rcises  on 
everv  subject.  The  book  possesses  many  attractions  for  the  teacher  and  stu- 
dent", and  is  destined  to  become  a  popular  school-book.  It  has  already  been 
introduced  into  many  of  the  principal  schools  and  colleges  in  the  country. 


f^EISSNER's     pERMAN     GrAMMAF^ 

A  Comparative  Eti{j7ls7i-Gertnan  Grammar,  ha9e<l  on  | 
the  aflanity  of  the  two  languages.    By  Prof.  Elias  Peissner, 
iate  of  the  University  of  '.lunich,  and  of  IJnion  College,  Sche- 
nectady.   New  edition,  revised.     316  pages. 


Tlie   Science   of    Govermnent   in    Connection'  irith 
American  Institutions,    By  Joseph  Alden,  D.D.,  LL.D., 
Pres.  of  State  Normal  School,  Albany.    1  vol  12mo. 
Adapted  to  the  wants  of  High  Schools  and  Colleges. 

Alden' s  Citizen's  Manual:  a  Text-Book  on  Government,  in 
Connection  witli  American  Institutions,  adapted  to  the  wants  of 
Common  Schools.  It  is  in  the  form  .of  questions  and  answers. 
By  Joseph  Alden,  D.D.,  LL.D,    1  vol.  16mo. 

Hereafter  no  American  can  be  said  to  be  educated  who  does  not  thoronghl- 
understand  the  formation  of  our  Government.  A  prominent  divine  has  t^aid. 
that  "  ever^  young  person  ehoukl  carefully  and  conscientiously  be  tauirlit  tb<>M- 
distinctive  ideas  which  constitute  the  substanceof  our  Constitution,  and  which 
determine  the  policy  of  our  politics  ;  and  to  this  end  there  ought  forthwith  to 
be  introduced  into  our  schools  a  simple,  comprehensive  manual,  whereby  the 
needed  tuition  should  be  implanted  at  that  early  period. 

JScJimitz's  3Ianual  of  Ancient  History;  from  the  Re- 
motest Times  to  the  Overthrow  of  the  Western  Empire.  A.  d. 
476,  with  copious  Chronological  Tables  and  Index.  By  Dr. 
Leoxhakd  Schmitz,  T.  Tl.  S.  K,  Edinburgh. 

The  Elements  of  Intellectual  Philosophy.  By  Fp.an  /? 
Watland,  D.D.    1  vol.  12ino. 

This  clearly-written  book,  from  the  pen  of  a  scholar  of  eminent  abil;(y,  mn 
who  has  had  the  largest  experience  in  the  education  of  '  .'  in  nnnil, 
id  unquestionably  at  the  head  of  text-books  in  Intellectual  PhilooopLy. 

An   Outline    of  the  Necessary  Laws   of    Thought: 

A  Treatise  on  Pure  and  Applied  Logic.  By  William  Thom- 
son, D.D.,  Provost  of  the  Queen's  College,  Oxford.  1  vol.  V^mo. 
Ooth. 

This  book  has  been  adopted  as  a  regular  text-book  in  Harvard,  Yale, 
Rochester,  New  York  University,  &c. 

Fairchilds'  Moral  Philosophy ;  or,  The  Science  of 
Obligation,  By  J.  H.  FAmcHiLD^,  Presi'dent  of  Oberlin 
College.    1  vol.  12mo. 

The  aim  of  this  volume  is  to  set  forth,  more  fully  than  has  hitherto  been 
done,  the  doctrine  that  virtue,  in  its  elementary  form,  consists  in  benevo- 
lence, and  that  all  forms  of  virtuous  action  are  modifications  of  this  principle. 

After  ,-resonting  this  view  of  obligation,  the  author  takes  up  the  questions  of 
Fiactical  Ethics,  Government  and  Personal  Rights  i*nd  Duties,  and  treat- 
them  in  their  relation  to  Benevolence,  aiming  at  a  solution  0(  the  problems  of 
rijjht  and  wrong  upon  ♦bis  simple  principle. 


:jn,v  c'  the  above  sent  biJ  maiZ,  post-paid,  on  receipt  of  price. 


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OLNEY'S    MATHEMATICAL.    SERIES. 


A  TREATISE 


ON 


Special  or  Elementary 


GEOMETKY. 


UNIVERSITY  EDITION. 


INCLUDING  AN  ELEMENTARY,   AND  ALSO,  IN  PART   III.,  A   HIGHER  COURSE,  IN 

PLANE,   SOLID,   AND   SPHERICAL   GEOMETRY;    AND    PLANE    AND    SPHERICAL 

TRIGONOMETRY,  WITH  THE  NECESSARY   TABLES. 


BY 

EDWAED    OLISTEY, 

PROFESSOR  OF  MATHEMATICS  IN  THE  UNIVERSITY  OF  MICHIQAN. 


NEW     YORK: 

Sheldon  &  Company, 

No.  8    MURRAY    STREET. 
1877. 


Stoddard's  lathematical  Series. 


STODDARD'S  JUVENILE  MENTAL  ARITHMETIC     - 
STODDARD'S  INTELLECTUAL  ARITHMETIC 
STODDARD'S  RUDIMENTS  OF  ARITHMETIC      - 
STODDARD'S  NEW  PRACTICAL  ARITHMETIC       - 

SHORT  AND  FULL  COURSE  FOR  GRADED  SCHOOLS. 

STODDARD'S  PICTORIAL  PRIMARY  ARITHMETIC 
STODDARD'S  COMBINATION  ARITHMETIC   .        .        -        . 
STODDARD'S  COMPLETE  ARITHMETIC     -        .        .        - 


Tlie  Combination  Scliool  Arithmetic  being  Mental  and  Written  Arithmetic 
in  one  book,  will  alone  serve  for  District  Schools.  For  Academies  a  full  high 
course  is  obtained  by  the  Complete  Arithmetic  and  InteUectiuil  Arithmetic. 


HIGHER     MATHEMATICS, 

BY 

EDWAED    OL:NrEY, 

TROPESSOR  OF  MATHEMATICS  IN  THE  UNIVERSITY  OP  MICHIGAN. 

A  COMPLETE  SCHOOL  ALGEBRA,  in  one  vol.,  410  pages,  Designed 

for  Elementary  and  higher  classes  in  Schools  and  Academies. 

A  (.GEOMETRY  AND  TRIGONOMETRY,  in  one  vol.,  8vo,       - 

A    GEOMETRY    AND    TRIGONOMETRY,    UNIVERSITY    EDITION,    in 
one  vol.,  8vo,    .        .        - 

A  GENERAL  GEOMETRY  AND  CALCULUS  in  one  vol.     - 

The  other  books  of  this  Series  will  be  published  as  rapidly  as  possible. 


Entered  according  to  Act  of  Congress,  in  the  year  1872,  by 

SHELDON  &  COMPANY. 
In  the  OfBce  of  the  Librarian  of  Congress,  at  Washington. 


^  /'^^  -J 


I  r 


PEEFACE.  '^"^^ 


This  treatise  on  the  Special  or  Elementary  Geometry  consists  of 
four  parts. 

Part  I.  is  designed  as  an  introduction.  In  it  the  student  is  made 
familiar  with  the  geometrical  concepts,  and  with  the  fundamental 
definitions  and  facts  of  the  science.  The  definitions  here  given,  are 
given  once  for  all.  It  is  thought  that  the  pupil  can  obtain  his  first 
conception  of  a  geometrical  fact,  as  well,  at  least,  from  a  correct, 
scientific  statement  of  it,  as  from  some  crude,  colloquial  form,  the 
language  of  which  he  will  be  obliged  to  replace  by  better,  after  the 
former  shall  have  become  so  firmly  fixed  in  his  mind,  as  not  to  be 
easily  eradicated.  No  attempt  at  demonstration  is  made  in  this  part, 
although  most  of  the  fundamental  facts  of  Elementary  Plane  Geom- 
etry are  here  presented,  and  amply  and  familiarly  illustrated.  This 
course  has  been  taken  in  obedience  to  the  canon  of  the  teacher's  art, 
which  prescribes  "  facts  before  theories."  Moreover,  such  has  been 
the  historic  order  of  development  of  this,  and  most  other  sciences; 
viz.,  the  facts  have  been  known,  or  conjectured,  long  before  men  have 
been  able  to  give  any  logical  account  of  them.  And  does  not  this 
indicate  what  may  he  the  natural  order  in  which  the  individual  mind 
will  receive  science  ?  When  the  student  has  become  familiar  with 
the  things  (coAcepts)  about  which  his  mind  is  to  be  occupied,  and 
knows  some  of  the  more  important  of  their  properties  and  relations, 
he  is  better  prepared  to  reason  upon  them. 

Part  II.  contains  all  the  essential  propositions  in  Plane,  Solid,  and 
Spherical  Geometry,  which  are  found  in  our  common  text-books,  with 
their  demonstrations.  The  subject  of  triedrals  and  the  doctrine  of 
the  sphere  are  treated  with  more  than  the  ordinary  fullness.  The 
earlier  sections  of  this  part  are  made  short,  each  treating  of  a  single 
subject,  and  the  propositions  are  made  to  stand  out  prominently.  At 
the  close  of  each  section  are  Exercises  designed  to  illustrate  and 
apply  the  principles  contained  in  the  section,  rather  than  to  extend 
the  pupil's  knowledge  of  geometrical  facts.  These  features,  together 
with  the  synopses  at  the  close  of  the  sections,  practical  teachers  can- 
not fail  to  appreciate. 

Part  III.,  which  is  contained  only  in  the  University  Edition,  has 


3V  •  PKEFACE. 

been  written  with  special  reference  to  the  needs  of  students  in  the 
University  of  Micliigan.  Our  admirable  system  of  public  High- 
Schools,  of  which  schools  there  is  now  one  in  almost  every  consid- 
erable village,  promises  ere  long  to  become  to  us  something  near 
what  the  German  Gymnasia  are  to  their  Universities.  In  order  to 
promote  the  legitimate  development  of  these  schools,  it  is  necessary 
that  the  University  resign  to  them  the  work  of  instruction  in  the 
elements  of  the  various  branches,  as  fast  and  as  far  as  they  are  pre- 
pared in  sufificient  numbers  to  undertake  it.  It  is  thought  that 
these  schools  should  now  give  the  instruction  in  Elementary  Geom- 
etry, which  has  hitherto  been  given  in  our  ordinary  college  course. 
The  first  two  parts  of  this  volume  furnish  this  amount  of  instruc- 
tion, and  students  are  expected  to  pass  examination  upon  it  on  their 
entrance  into  the  University.  This  amount  of  preparation  enables 
students  to  extend  their  knowledge  of  Geometry,  during  the  Fresh- 
man year  in  the  University,  considerably  beyond  what  has  hitherto 
been  practicable.  As  a  text-book  for  such  students.  Part  III.  has 
been  written.  At  this  stage  of  his  progress,  the  student  is  prepared 
to  learn  to  investigate  for  himself.  Hence  he  is  here  furnished  with 
a  large  collection  of  well  classified  theorems  and  problems,  which 
afford  a  review  of  all  that  has  gone  before,  extend  his  knowledge  of 
geometrical  truth,  and  give  him  the  needed  discipline  in  original 
demonstration.  To  develop  the  power  of  independent  thought,  is 
the  most  difficult,  while  it  is  the  most  important  part  of  the  teach- 
er's work.  Great  pains  have  therefore  been  taken,  in  this  part 
of  the  work,  to  render  such  aid,  and  only  such,  as  a  student  ought  to 
require  in  advancing  from  the  stage  in  which  he  has  been  follow- 
ing the  processes  of  others,  to  that  of  independent  reasoning.  In 
the  second  place,  this  part  contains  what  is  usually  styled  Applica- 
tions  of  Algebra  to  Geometry,  with  an  extended  and  carefully  selected 
range  of  examples  in  this  important  subject.  A  third  purpose  has 
been  to  present  in  this  part  an  introduction  to  what  is  often  spoken 
of  as  the  Modern  Geometry,  by  which  is  meant  the  results  of  modern 
thought  in  developing  geometrical  truth  upon  the  direct  method. 
While,  as  a  system  of  geometrical  reasoning,  this  Geometry  is  not 
philosophically  different  from  that  with  which  the  student  of  Euclid 
is  familiar,  and  which  is  properly  distinguished  as  the  special  or  direct 
method,  the  character  of  the  facts  developed  is  quite  novel.  So 
much  so,  indeed,  that  the  student  who  has  no  knowledge  of  Geometry 
but  that  which  our  common  text-books  furnish,  knows  absolutely 
nothing  of  the  domain  into  which  most  of  the  brilliant  advances  of 


PREFACE.  V 

the  present  century  have  been  made.  He  knows  not  even  the  terms 
in  which  the  ideas  of  such  w^riters  as  Poncelet,  Chasles,  and  Sal- 
mon, are  expressed,  and  he  is  quite  as  much  a  stranger  to  the  thought. 
In  this  part  are  presented  the  fundamental  ideas  concerning  Lociy 
Symmetry,  Maxima  a)id  Minima,  Isox>erimeiry,  the  theory  of  Trans- 
versals, Anharmonic  Ratio,  Polars,  Radical  Axes,  and  other  modern 
views  concerning  the  circle. 

Part  IV.  is  Plane  and  Spherical  Trigonometry,  with  the  requisite 
Tables.  While  this  Part,  as  a  whole,  is  much  more  complete  than  the 
treatises  in  common  use  in  our  schools,  it  is  so  arranged  that  a  shorter 
course  can  be  taken  by  such  as  desire  it.  Thus,  for  a  shorter  course  in 
Plane  Trigonometry,  see  Note  on  page  55.  In  Spherical  Trigono- 
metry, the  first  three  sections,  either  with  or  without  the  Introduc- 
tion on  Projection,  will  afford  a  very  satisfactory  elementary  course. 

A  few  words  as  to  the  manner  in  which  this  plan  has  been  executed, 
may  be  important.  In  general,  the  Definitions  are  those  usually  given, 
with  such  slight  alterations  as  have  been  suggested  by  reflection  and 
experience.  There  are,  however,  a  few  exceptions.  Among  these  is 
the  definition  of  an  Angle.  I  can  but  regard  the  attempt  to  define 
an  angle  as  The  differ eiice  in  direction  hetiueen  tiuo  lines,  or  The 
amount  of  divergence,  as  needlessly  vague,  abstract,  and  perplexing 
to  a  student,  as  well  as  questionable  on  philosophical  grounds.  The 
definition  given  in  the  text  will  be  seen  to  be,  at  bottom,  the  old 
one,  the  conception  being  slightly  altered  to  bring  it  into  more  close 
connection  with  common  thought,  and  also  with  the  idea  of  an  angle 
as  generated  by  the  revolution  of  a  line.  As  to  Parallels,  and  the 
definition  of  similarity,  my  experience  as  a  teacher  is  decidedly  in 
favor  of  retaining  the  old  notions.  So  also  in  adopting  a  definition 
of  a  Trigonometrical  Function,  I  am  compelled  to  adhere  to  the 
geometrical  conception.  A  ratio  is  a  complex  concept,  and  conse- 
quently not  so  easy  of  application  as  a  simple  one.  For  this  reason, 
among  others,  I  prefer  the  differential  to  the  differential  coefficient, 
in  the  calculus,  and  a  line  to  a  ratio,  in  Trigonometry.  Moreover,  I 
have  found  that  students  invariably  rely  upon  the  geometrical  con-^ 
ception,  even  when  first  taught  the  other ;  hence  I  am  not  surprised- 
that  all  our  writers  who  define  a  trigonometrical  function  as  a  ratio, 
hasten  to  tell  the  pupil  what  it  means,  by  giving  him  the  geometrical 
illustrations.  Nor  are  the  superior  facility  which  the  geometrical 
conception  affords  for  a  full  elucidation  of  the  doctrine  of  the  signs 
of  the  functions,  and  its  admirable  adaptation  to  fix  these  laws  in  the 
mind,  considerations  to  be  lost  sight  of  iu  selecting  the  definition. 


▼1  PREFACE. 

Surely  no  apology  is  needed,  at  the  present  day,  for  introducing 
the  idea  of  motion  into  Elementary  Geometry,  notwithstanding  the 
rigorous  and  disdainful  manner  with  which  its  entrance  was  long  re- 
sisted by  the  old  Geometers.  And,  having  admitted  this  idea,  the 
conception  of  loci  as  generated  by  motion  would  seem  to  follow  as  a 
logical  necessity.  In  like  manner,  I  take  it,  the  Infinitesimal 
method  must  come  in.  Its  directness,  simplicity,  and  necessity  in 
applied  mathematics,  demand  its  recognition  in  the  elements.  In 
two  or  three  instances,  I  have  presented  the  reductio  ad  ahsicrdum, 
where  the  methods  are  equivalents,  and  have  always  in  presenting  the 
infinitesimal  method  woven  in  the  idea  of  limits,  which  I  conceive  to 
be  fundamentally  the  same  as  the  infinitesimal.  Thus  we  bring  the 
lower  and  higher  mathematics  into  closer  connection. 

The  order  of  arrangement  in  Plane  Geometry  (Chap.  L),  is  thought 
to  be  simple,  philosophical,  and  practical.  A  glance  at  the  table  of 
contents  will  show  what  it  is.  This  arrangement  secures  the 
very  important  result,  that  each  section  presents  some  particulai 
method  of  2^roof  and  holds  the  student  to  it,  until  it  is  familiar. 
True,  it  requires  that  a  larger  number  of  propositions  be  demonstrated 
from  fundamental  truths ;  but  who  will  consider  this  an  objection  ? 

To  such  as  consider  it  the  sole  province  of  geometrical  demonstra- 
tion, to  convince  the  mind  of  the  truth  of  a  proposition,  not  a  few 
theorems  in  these  and  ordinary  pages  must  seem  quite  superfluous.  To 
them.  Prop.  I.,  page  121,  may  afford  some  merriment.  But  those  who, 
with  myself,  consider  Geometry  as  a  branch  of  practical  logic,  the 
aim  of  which  is  to  detect  and  state  the  steps  which  actually  lie  be- 
tween premise  and  conclusion,  will  see  the  propriety  of  such  demonstra- 
tions; and  for  each  individual  of  the  other  class,  a  separate  treatise 
will  be  needed,  since  no  two  minds  will  intuitively  grant  exactly  the 
same  propositions. 

To  Ex-President  Hill,  of  Harvard,  I  am  indebted  for  the  confir- 
mation of  an  opinion  which  had  been  previously  forming  in  my  mind, 
that  the  study  of  Geometry  as  a  branch  of  logic,  should  be  preceded 
by  a  presentation  of  its  leading  facts.  The  works  of  Compagxon", 
Tappan,  and  our  lamented  countryman,  Chauyexet,  have  been 
within  reach  during  the  entire  work  of  preparation,  and  this  volume 
would  have  been  different,  in  some  respects,  if  any  one  of  these  able 
treatises  had  not  aj^peared  before  it. 

In  the  preparation  of  Part  III.  the  works  of  Rouche  et  Combe- 
ROUSSE  and  Mulcahy  have  been  freely  used.  For  the  very  concise 
and  elegant  form  in  which  the  principle  of  Delambre,  for  the  pre- 


PREFACE.  Vil 

cise  calculations  of  Trigonometrical  Functions  near  their  limits,  is 
embodied  in  Table  III.,  I  am  indebted  to  the  recent  work  of  Presi- 
dent Eli  T.  Tappan,  of  Kenyon  College,  Ohio. 

My  long  and  intimate  intercourse  with  Professor  G.  B.  Merriman", 
now  of  the  department  of  Physics  in  the  University,  has  been  a 
source  of  great  profit  to  me  in  the  preparation  of  the  entire  work. 
His  sound,  practical  judgment  as  a  teacher  of  Geometry,  and  culti- 
vated taste  and  skill  as  a  Mathematician,  have  been  ever  at  my  ser- 
vice, and  have  done  more  than  I  can  tell,  in  giving  form  to  the  work, 
both  as  respects  its  matter  and  its  spirit. 

Edward  Olney. 

University  of  Michigan, 

Ann  Arbor,  January,  1872. 


COFTEWTS. 


INTRODUCTION". 


Paos. 
SECTION  I. 
Lo(jico-Mathematical  Terms  Defined  and  Illustrated 1-3 

SECTION  II. 
The  Geometrical  Concepts  Defined  and  Illustrated 3-11 


PART  L 

A  FEW  OF  THE  MORE  IMPORTANT  FACTS  OF  THE  SCIENCE. 

SECTION  I. 
About  Straight  Lines 11-19 

SECTION  11. 
About  Circles 19-25 

SECTION  III. 
About  Angles  and  Parallels 25-30 

SECTION  IV. 
About  Triangles 31-35 

SECTION  V. 
About  Equal  Figures 35-39 

SECTION  VI. 
About  Similar  Figxhies,  especially  Triangles 39-44 

SECTION  VIL 
About  Areas 44-55 

SECTION   VIII. 
About  Polygons 55-58 


X  CONTENTS. 

PART  11.  • 

THE     FUNDAMENTAL     PROPOSITIONS     OF    ELEMENTARY     GEOMETRY, 
DEMONSTRATED,   ILLUSTRATED,   AND   APPLIED. 


CHAPTER  I. 
PLANE  GEOMETRY. 

SECTION  I.                                   >  Page. 

Pkrpendicular  Straight  Lines 60-65 

SECTION  II. 
Oblique  Straight  Lines 65-70 

SECTION  III. 
Parallels 70-77 

SECTION  IV. 
Relative  Positions  op  Straight  Lines  and  Circumferences 7&-85 

SECTION  V. 
Relative  Positions  of  Circidiferences 80-93 

SECTION  VI. 
Measurement  of  Angles 94-103 

SECTION  vn. 

Angles  of  Polygons  and  the  Relation  between  the  Angles  and  Sides. 

Of  Triangles 104-106 

Of  Quadi-ilaterals 107-111 

Of  Polygons  of  more  than  Four  Sides 112-113 

Of  Regular  Polygons 113-120 

SECTION  vni. 

Of  Equality. 

Of  Lines  and  Circles 121-122 

Of  Angles 122-123 

Of  Triangles 124-129 

Of  Quadrilaterals 129-130 

Of  Polygons  of  more  than  Four  Sides 130-137 

SECTION  IX. 
Op  Equivalency  and  Areas. 

Equivalency 138-140 

Area 140-144 


CONTENTS.  ti 

SECTION  X.  Paor. 

Of  Similakity 144-152 

SECTION  XL 

Applications  of  the  Doctrine  op  Sdhlarity  to  the  Development  op 
Geometrical  Properties  of  Figures. 

Of  the  Relations  of  the  Segments  of  two  Lines  intersecting  each 

other  and  intersected  by  a  circumference 153-154 

Of  the  Bisector  of  an  Angle  of  a  Triangle 154-156 

Areas  of  Similar  Figures 156-158 

Perimeters  and  the  Rectification  of  the  Circumference 158-160 

Area  of  the  Cucle 160-163 


CHAPTER  IL 
SOLID  GEOMETRY. 

SECTION  I. 
Of  Straight  Lines  and  Planes.  Paob. 

Perpendicular  and  Obhque  Lines  to  a  Plane 164-169 

Parallel  Lines  and  Planes 169-174 

SECTION  IL 
Of  Solid  Angles. 

Of  Diedrals 176-178 

OfTriedrals 178-184 

Of  Polyedrals 185-186 

SECTION  IIL 
Op  Prisms  and  Cylinders 187-199 

SECTION  IV. 
Of  Pyramids  and  Cones 199-209 

SECTION  V. 
Of  the  Sphere. 

Circles  of  the  Sphere 210-211 

Distances  on  the  Surface  of  a  Sphere 211-215 

Spherical  Angles 215-218 

Tangent  Planes 218-219 

Spherical  Triangles 219-226 

Polar  or  Supplemental  Triangles 226-228 

Quadrature  of  the  Surface  of  the  Sphere 229-231 

Lunes 231-235 

Yoiume  of  Sphere 235-239 


XU  CONTENTS. 

PART  IIL 

AN^  ADVANCED   COURSE   IN*   GEOMETRY. 

CHAPTER  I. 
EXERCISES  IN  GEOMETRICAL  INVENTION. 

SECTION   I.  Page. 

Theorems  m  Special  or  Elementary  Geometry 243-267 

SECTION  II. 

Problems  in  Special  or  Elementary  Geometry 267-276 

SECTION  III. 
Applications  op  Algebra  to  Geometry 276-288 

CHAPTER   11. 

INTRODUCTION  TO  MODERN  GEOMETRY. 

SECTION  I. 

Of  Loci 288-296 

SECTION  11. 

Op  Symmetry 296-30 1 

SECTION  III. 

Of  Maxima  and  Minima,  and  Isoperimetry 302-306 

SECTION  IV. 

Of  Trans\t:rsals  30&-310 

SECTION  V. 

Harmonic  Proportion,  and  Harmonic  Pencils 310-314 

SECTION  VI. 

Anharmonic  Ratio 314-318 

SECTION  VII. 

Pols  and  Polar  in  Respect  to  a  Circle 319-323 

SECTION  VIII. 
Radical  Axes  and  Centres  op  Similitude  op  Circles 323-329 


SPECIAL  OR  ELEMENTARY 

GEOMETRY. 


INTBOD  UCTION. 


SECTION  I, 

LOGICO-MATHEMATICAL  TERMS.* 

JZ,  a  ^Proposition  is  a  statement  of  something  to  be  considered 
or  done. 

III. — Thus,  the  common  statement,  "  Life  is  short,"  is  a  proposition ;  so, 
also,  we  make,  or  state  a  proposition,  when  we  say,  '*  Let  us  seek  earnestly  after 
truth."—"  The  product  of  the  divisor  and  quotient,  plus  the  remainder,  equals 
the  dividend,"  and  the  requirement,  "  To  reduce  a  fraction  to  its  lowest  terms," 
are  examples  of  Arithmetical  propositions. 

2.  Propositions  are  distinguished  as  Axioms,  Theorems,  Lemmas, 
Corollaries,  Postulates,  and  ProUems. 

3.  An  Axiom  is  a  proposition  which  states  a  principle  that 
is  so  simple,  elementary,  and  evident  as  to  require  no  proof. 

Ill,— Thus,"  A  part  of  a  thing  is  less  than  the  whole  of  it,"  "  Equimultiples 
of  equals  are  equal,"  are  examples  of  axioms.  If  any  one  does  not  admit  the 
Iruth  of  axioms,  when  he  understands  the  terms  used,  we  say  that  his  mind  is 
not  sound,  and  that  we  cannot  reason  with  him. 

4.  A.  Theorem  is  a  proposition  which  states  a  real  or  supposed 
fact,  whose  truth  or  falsity  we  are  to  determine  by  reasoning. 

III. — '*  If  the  same  quantity  be  added  to  both  numerator  and  denominator 
of  a  proper  fraction,  the  value  of  the  fraction  will  be  increased,"  is  a  tJieoi-em. 
It  is  a  statement  the  truth  or  falsity  of  which  we  are  to  determine  by  a  course 
of  reasoniH^. 

*  That  i?,  terms  used  in  the  science  in  consequence  of  its  logical  character.  The  science  of. 
the  Pure  Mathematics  may  be  considered  as  a  department  of  practical  logic. 


2  LOGICO-MATHEMATICAL  TERMS. 

5,  A  T>enionstvation  is  the  course  of  reasoning  by  means 
of  which  the  truth  or  falsity  of  a  theorem  is  made  to  appear.  The 
term  is  also  applied  to  a  logical  statement  of  the  reasons  for  the 
processes  of  a  rule.  A  solution  tells  lioiv  a  thing  is  done:  a  demon- 
stration tells  vjhy  it  is  so  done.  A  demonstration  is  often  called 
proof. 

6,  A  Lemma  is  a  theorem  demonstrated  for  the  pui-pose  of 
using  it  in  the  demonstration  of  another  theorem. 

III. — Thus,  in  order  to  demonstrate  the  rule  for  finding  the  greatest  common 
divisor  of  two  or  more  numbers,  it  may  be  best  first  to  prove  that  "  A  divisor 
of  two  numbers  is  a  divisor  of  their  sum,  and  also  of  their  difi'erence."  This 
theorem,  when  proved  for  such  a  purpose,  is  called  a  Lemma. 

The  term  Lemma  is  not  much  used,  and  is  not  very  important,  since  most 
theorems,  once  proved,  become  in  turn  auxiliary  to  the  proof  of  others,  and 
hence  might  be  called  lemmas. 

7,  A  Corollary  is  a  subordinate  theorem  which  is  suggested, 
or  the  truth  of  which  is  made  evident,  in  the  course  of  the  demon- 
stration of  a  more  general  theorem,  or  which  is  a  direct  inference 
from  a  proposition,  or  a  definition. 

III. — Thus,  by  the  discussion  of  the  ordinaiy  process  of  performing  subtrac- 
tion in  Arithmetic,  the  following  Corollary  m\g\i\.  be  suggested:  "Subtraction 
may  also  be  performed  by  addition,  as  we  can  readily  obseiTe  what  number 
must  be  added  to  the  subtrahend  to  produce  the  minuend." 

8,  A  JPostiilate  is  a  proposition  which  states  that  something 
can  be  done,  and  which  is  so  evidently  true  as  to  require  no  process 
of  reasoning  to  show  that  it  is  possible  to  be  done.  We  may  or  may 
not  know  how  to  perform  the  operation. 

III. — Quantities  of  the  same  kind  can  be  added  together. 

0,  A  JProblem  is  a  proposition  to  do  some  specified  thing,  and 
is  stated  with  reference  to  developing  the  method  of  doing  it. 

Ill.^-A  problem  is  often  stated  as  an  incomplete  sentence,  as,  "  To  reduce 
fractions  to  a  common  denominator." — This  incomplete  statement  means  that 
"  "We  propose  to  show  how  to  reduce  fractions  to  a  common  denominator." 
Again,  the  problem  "  To  construct  a  square,"  means  that  "  We  propose  to  draw 
a  figure  which  is  called  a  square,  and  to  tell  how  it  is  done." 

10,  A  Utile  is  a  formal  statement  of  the  method  of  solving  a 
general  problem,  and  is  designed  for  practical  application  in  solving 
special  examples  of  the  same  class.  Of  course  a  rule  requires  a 
demonstration. 


THE   GEOMETRICAL   CONCEPTS. 


11.  A  Solution  is  the  process  of  performing  a  problem  or  an 
example.  It  should  usually  be  accompanied  by  a  demonstration  of 
the  process. 

12,  A  Scholium  is  a  remark  made  at  the  close  of  a  discussion, 
and  designed  to  call  attention  to  some  particular  feature  or  features 
of  it. 

III. — Thus,  after  having  discussed  the  subject  of  multiplication  and  division 
In  Arithmetic,  the  remark  that  "  Division  is  the  converse  of  multiplication,"  is 
a  scholium. 


SYNOPSIS. 

Subject  of  the  section. 

Lemma. 

m.    Why  the 

Proposition.    HI. 

portant. 

Varieties  of  propositions. 

Corollary. 

HI. 

Axiom.    ML 

Postulate. 

El. 

One  who  will  not  admit  the  truth 

Problem. 

How  stated. 

of  axioms. 

Rule. 

Theorem.     III. 

Solution. 

Demonstration.   Difference  between 

Scholium. 

m. 

a  solution  and  a  demonstration. 

m. 


SECTION  11. 

THE  GEOxMETRICAL  CONCEPTS.* 


FOUNTS. 

IS.  A  JPoint  is  a  place  without  size, 
letters. 


Points  are  designated  by 


•B 


III. — If  we  wish  to  designate  any  particular  point  (place)  on  the  paper,  we 
put  a  letter  by  it,  and  sometimes  a  dot  on  it.  Thus, 
in  Fig.  3,  the  ends  of  the  line,  which  are  points,  are  ^p 

flesignated  as  "  point  A,"  "  point  D  ;"  or,  simply, 
as  A  and  D.    The  points  marked  on  the  line  are 

designated  as  "  point  B,"  "  point  C,"  or  as  B  and    a B — 

C.     F  and  E  are  two  points  above  the  line.  Fig. 


*  A  concept  is  a  thing  thought  about ;— a  thought-object.  Thus,  in  Arithmetic,  number  is 
tbe  concept ;  in  Botany,  plants ;  in  Geometry,  as  will  appear  in  this  section,  pointP,  lines,  and 
solids.    These  may  aleo  be  said  to  conetitate  the  suloect-matter  of  the  science. 


ELEMENTARY  GEOMETRY. 


LINES. 

14,  A  lAne  is  the  path  of  a  point  in  motion.  Lines  are  repre- 
sented upon  paper  by  marks  made  with  a  pen  or  pencil,  the  point  of 
the  pen  or  pencil  representing  the  moving  point.  A  line  is  desig- 
nated by  naming  the  letters  written  at  its  extremities,  or  somewhere 
upon  it. 

III. — In  each  case  in  Fig.  4,  conceive  a  point  to  start  from  A  and  move  along 
^  the  path  indicated  by  the  mark  to  B.    The  path 

thus  traced  is  a  line.  Since  a  true  point  has  no 
size,  a  line  has  no  breadth,  though  the  marks  by 
•which  we  represent  lines  have  some  breadth. 
The  first  and  third  lines  in  the  figure  are  each 
designated  as  "the  line  AB."  The  second  line 
is  considered  as  traced  hy  a  point  starting  from 
A  and  coming  around  to  A  again,  so  that  B  and  A 
coincide.  This  line  may  be  designated  as  the 
line  AmnA,  or  A7?i?iB.  In  the  fourth  case,  there 
are  three  lines  represented,  which  are  designated, 
respectively,  as  AwiB,  A;iB,  and  Ac3 ;  or,  the 
last,  as  AB. 

15.  Lines    are    of   Two    Kiiuls, 

Straight  and  Curved.  A  straight  line  is 
also  called  a  EigJit  Line.  A  curved  line 
is  often  called  simply  a  Curve. 

16,  A  Straiffht  Line*  is  a  Une 

traced  by  a  point  which  moves  constantly 
in  the  same  direction. 


17.  A  Curved  Line  is  a  line  traced  by  a  point  which  con- 
stantly changes  its  direction  of  motion. 

Ill's.— Thus  in  1,  Fig.  4,  if  the  line  AB  is  conceived  as  traced  by  a  point 
moving  from  A  to  B,  it  is  evident  that  this  point  moves  in  the  same  direction 
throughout  its  course;  hence  AB  is  a  straight  line.  If  a  body,  as  a  stone,  be 
let  fall,  it  moves  constantly  toward  the  centre  of  the  earth ;  hence  its  path 
represents  a  straight  line.  If  a  weight  be  suspended  by  a  string,  the  string 
represents  a  straight  line.  Considering  the  line  represented  by  AiB,  Fig.  4,  as 
the  path  of  a  point  moving  from  A  to  B,  we  see  that  the  direction  of  motion  is 
constantly  changuig.    For  example,  if  this  were  a  line  traced  on  a  map,  we 


♦  The  word  ''line"  u*cd  aloue  Pi^'niSe;?  '-straight  line. 


SURFACES.  5 

would  say,  that,  starting  from  A,  the  point  begins  to  move  nearly  north,  but 
keeps  changing  its  direction  more  and  more  toward  the  east,  until  at  3  it  moves 
directly  east;  and  from  3  it  continues  to  change  its  course  and  moves 
more  and  more  toward  the  south,  till  at  i  it  is  moving  directly  south.  The 
same  general  ti-uth  is  illustrated  in  2  and  4,  Fig.  4.  The  path  of  a  ball  thrown 
into  the  air,  in  any  direction  except  directly  up,  represents  a  curved  line.  Most 
of  the  lines  seen  in  nature  are  curved,  as  ^ 

the  edges  of  leaves,  the  shore  of  a  river  7/^- 

or  lake,  etc.    Sometimes  a  path  like  that       ..^-^-^""^   X.^ 
represented  in  Fig.  5  is  called,  though  im-     ^  rv        o 

properly,  a  Broken  Line.    It  is  not  a  line  ^^®-  ^• 

at  all ;  that  is,  not  one  line  :  it  is  a  series  of  straight  lines. 


SURFACES. 
IS.  A  Surface  is  the  path  of  a  line  in  motion.* 
1.0*  Surfaces  are  of  Two  Kinds^  Plane  and  Curved. 

20.  A  I^lane  Surface,  or  simply  a  Plane,  is  a  surface  with 
which  a  straight  line  may  be  made  to  coincide  in  any  direction. 
Such  a  surface  may  always  be  conceived  as  the  path  of  a  straight 
line  in  motion. 

21.  A  Curved  Surface  is  a  surface  in  which,  if  lines  are 
conceived  to  be  drawn  in  all  directions,  some  or  all  of  them  will 
be  curved  lines. 

Ill's. — Let  AB,  Fig.  6,  be  supposed  to  move  to  the  right,  so  that  its  extremi- 
ties A  and  B  move  at  the  same  rate  and  in  the 
same  direction,  A  tracing  the  line  AD,  and  B,  the 
line  BC.  The  path  of  the  line,  the  figure  ABCD, 
,is  a  surface.  This  page  is  a  surface,  and  may  be 
conceived  as  the  path  of  a  line  sliding  like  a  ruler 
from  top  to  bottom  of  it,  or  from  one  side  to  the 
other.  Such  a  path  will  have  length  and  breadth, 
being  in  the  latter  respect  unlike  a  line,  which  has  Fig.  6. 

only  length. 

As  a  second  illustration,  suppose  a  fine  wire  bent  into  the  form  of  the 
curve  AmB,  Fig.  7,  and  its  ends  A  and  B  stuck  into  a  rod,  XY.  Now,  taking  the 
rod  XY  in  the  fingers  and  rolling  it,  it  is  evident  that  the  path  of  the  hne 
represented  by  the  wire  AwB,  will  be  the  surfoce  of  a  ball  (sphere). 

♦  Should  it  be  paid  that  irrefjular  surfaces  are  not  included  in  this  definition,  the  sufflcient 
reply  is,  that  such  surfaces  are  not  subjects  of  Geometrical  investigation,  except  approzi* 
mately,  by  means  of  regular  surfaces. 


6 


ELEMENTARY  GEOMETRY. 


nv 


X  A 


Fig. 


B  Y 


Again,  suppose   the  rod  XY  be  placed  on  the  surface  of  this  paper  so 

that  the  wire  AmB  shall  stand  straight  up 
from  the  paper,  just  as  it  would  be  if  we 
could  take  hold  of  the  curve  at  m  and  raise 
it  right  up,  letting  XY  lie  as  it  does  in  the 
figure.  Now  slide  the  rod  straight  up  or 
down  the  page,  making  both  ends  move  at  the 
same  rate.  The  path  of  AwiB  will  be  like  tlie 
smface  of  a  half-round  rod  (a  semi-cylinder). 

Thus  we  see  how  surfaces  plane  and  cuiTed  may  be  conceived  as  the  paths  of 

lines  in  motion. 

^^  Ex.  1.  If  the  curve  A;iB,  Fuj.  8, 

be  conceived  as   revolved   about 
the  line  XY,  the  surface  of  what 
Y      object  will  its  path  be  like  ? 


X  A 


Fig.  S. 


Ex.  2.  If  the  figure  OMNP,  Fig.  9,  be 
conceived  as  revolved  about  OP,  what  kind 
of  a  path  will  MN  trace?  What  kind  of 
paths  will  FN  and  OM  trace? 

Ans.  One  path  will  be  like  the  surface 
of  a  joint  of  stove-pipe,  i.  e.,  a  cylindrical 
surface ;  and  one  "will  be  like  a  flat  wheel, 
i,  e.,  a  circle. 


Ex.  B.  If  you  fasten  one  end  of  a  cord  at  a  point  in  the  ceiling  and 
hang  a  ball  on  the  other  end,  and  then  make  the  ball  swing  around 
in  a  circle,  what  kind  of  a  surface  will  the  string  describe  ? 

[Note. — The  student  is  not  necessarily  expected  to  give  the  geometrical 
name  of  the  surface,  but  rather  to  tell  in  his  own  way  what  it  is  like,  so  as  to 
make  it  clear  that  he  conceives  the  thing  itself] 

Ex.  4.  If  you  were  to  draw  lines  in  all  directions  on  the  surface  of 
the  stove-pipe,  might  any  of  them  be  straight  ?  Could  all  of  them 
be  straight  ?     What  kind  of  a  surface  is  this,  therefore  ? 

Ex.  5.  Can  you  draw  a  straight  line  on  the  surface  of  a  ball  ?  On 
the  surface  of  an  egg  ?    What  kind  of  surfaces  are  these  ? 

Ex.  6.  When  the  carpenter  wishes  to  make  the  surface  of  a  board 
perfectly  flat,  he  takes  a  ruler  whose  edge  is  a  straight  line,  and  lays 
this  straight  edge  on  the  surface  in  all  directions,  watching  closely 


ANGLES.  7 

to  see  if  ir,  always  touches.     Which  of  our  definitions  is  he  illus- 
trating by  his  practice  ? 

Ex.  7.  When  the  miller  wishes  to  make  flat  the  surface  of  one  of 
the  large  stones  with  which  wheat  is  ground  into  flour,  he  sometimes 
takes  a  ruler  with  a  straight  edge,  and  smearing  the  edge  with  paint, 
applies  it  in  all  directions  to  the  surface,  and  then  chips  off  the  stone 
where  the  paint  is  left  on  it.    AVhat  principles  is  he  illustrating  ? 

Ex.  8.  How  can  you  conceive  a  straight  line  to  move  so  that  it 
shall  not  generate  a  surface  ? 


ANGLES. 

22.  A  JPlane  Angle ^  or  simply  an  A7igle,  is  the  opening  be- 
tween two  lines  which  meet  each  other.  The  point  in  which  the 
lines  meet  is  called  the  vertex,  and  the  lines  are  called  the  sides. 
An  angle  is  designated  by  placing  a  letter  at  its  vertex,  and  one  at 
each  of  its  sides.  In  reading,  we  name  the  letter  at  the  vertex  when 
there  is  but  one  vertex  at  the  point,  and  the  three  letters  when  there 
are  two  or  more  vertices  at  the  same  point.  In  the  latter  case,  the 
letter  at  the  vertex  is  put  between  the  other  two. 

III. — In  common  language  an 
angle  is  called  a  corner.  The 
opening  between  the  two  lines 
AB  and  AC,  in  wiiich  the  figure  1 
stands,  is  called  the  angle  A  ;  or, 
if  we  choose,  we  may  call  it  the 
angle  BAC.  At  L  there  are  two 
vertices,  so  that  were  we  to  say 
the  angle  L,  one  would  not  know 
whether  we  meant  the  angle  (cor- 
ner) in  which  4  stands,  or  that  in 
which  5  stands.  To  avoid  this 
ambiguity,  we  say  the  angle  HLR 
for  the  former,  and  RLT  for  the 
latter.  The  angle  ZAY  is  the  cor- 
ner in  which  11  stands ;  that  is, 
the  opening  between  the  two 
lines  AY  and  AZ.  In  designating 
an  angle  by  three  lettere,  it  is  im- 
material which  letter  stands  first 
so  that  the  one  at  the  vertex  is 
put  between  the  other  two.  Thus, 
PQS  and  SQP  are  both  designa- 
tions of  the  angle  in  which  G 


Fig.  10. 


8  ELEMENTARY  GEOMETRY. 

stands.    An  angle  is  also  frequently  designated  by  putting  a  letter  or  figure  in 
it  and  near  the  vertex. 

23,  The  Size  of  an  Angle  depends  upon  the  rapidity  with 
which  its  sides  separate,  and  not  upon  their  length. 

III. — The  angles  BAG  and  MON,  Fig.  10,  are  equal,  since  the  sides  separate 
at  the  same  rate,  although  the  sides  of  the  latter  are  more  prolonged  than  those 
of  the  former.  The  sides  DF  and  DE  separate  faster  than  AB  and  AC,  hence  the 
angle  EDF  is  greater  than  the  angle  BAG. 

24:,  Adjacent  Angles  are  angles  so  situated  as  to  have  a  com- 
mon vertex  and  one  common  side  lying  between  them. 

III. — In  Fig.  10,  angles  4  and  5  are  adjacent,  since  they  have  the  common 
vertex  L,  and  the  common  side  LR.  Angles  9  and  10  are  also  adjacent,  as  are 
also  8  and  9. 

25.  Angles  are  distinguished  as  Right  Angles  and  Oblique  Angles, 
Oblique  angles  are  either  Acute  or  Obtuse. 

26,  A  Itight  Angle  is  an  angle  included  between  two  straight 
lines  which  meet  each  other  in  such  a  manner  as  to  make  the  adja- 
cent angles  equal.  An  Acute  Angle  is  an  angle  which  is  less 
than  a  right  angle,  i.  e.,  one  whose  sides  separate  less  rapidly. 
An  Obtuse  Angle  is  an  angle  which  is  greater  than  a  right  angle, 
i,  e.,  one  whose  sides  separate  more  rapidly. 

B  III. — As  in  common  language  an  angle  is  called 

a  corner,  so  a  right  angle  is  called  a  square  corner ; 
an  acute,  a  «/i«rp  corner;  and  an  obtuse  angle  might 
be  called  a  blunt  corner.  In  Fig.  11,  BAG  and 
DAB  are  right  angles.  In  Fig.  10,  1,  2,  3,  5,  8,  9, 
and  10  are  acute  angles,  4  and  6  are  obtuse,  and  7  is 
a  right  angle. 


A 
Fig. 


A  SOLID. 


27*  A  Solid  is  a  limited  portion  of  space.    It  may  also  be  con- 
ceived as  the  path  of  a  surface  in  motion. 

III. — Suppose  you  have  a  block  of  wood  like  that  represented  in  Fig.  12, 
with  all  its  corners  (angles)  square  corners  (right  angles).    Hold  it  still  in  your 


A   SOLID. 


fingers  a  moment,  and  fix  your  mind 
upon  it  .  Now  take  the  block  away  and 
think  of  the  space  (place)  where  it  was. 
This  space  will  be  of  just  the  same  form 
as  the  block  of  wood,  and  by  a  little  ef- 
fort you  can  think  of  it  just  as  well  as  of 
the  wood.  This  space  is  an  example  of 
what  we  call  a  Solid  in  Geometiy.  In 
fact,  the  solids  of  Geometry  are  not  solids  ^i^- 1^- 

at  all  in  the  common  sense  of  solids ;  they  are  only  just  places  of  certain  shapes. 

Again,  hold  your  ball  still  a  moment  in  your  fingers  and  then  let  it  drop,  and 
think  of  the  place  it  filled  when  you  had  it  in  your  fingers.  It  is  this  place, 
shaped  just  like  your  ball,  that  we  think  about,  and  talk  about  as  a  solid,  in 
Geometry. 

In  order  to  see  how  a  solid  may  be  conceived  as  the  path  of  a  surface,  sup- 
pose you  cut  out  a  piece  of  paper  of  just  the  same  size  as  the  end  of  the  block 
represented  in  Fig.  12.  Let  ABCD  represent  this  piece  of  paper.  Now,  holding 
the  paper  in  a  perpendicular  position,  as  ABCD  is  represented  in  the  figure, 
move  it  along  to  the  right,  so  that  its  angles  shall  trace  the  lines  AC,  BH,  DE, 
and  CF.  When  the  paper  has  moved  to  the  position  CHFE,  its  path  will  be 
just  the  same  space  as  the  block  of  wood  occupied.  This  path,  or  the  space 
through  which  the  surface  represented  by  the  piece  of  paper  moved,  is  the  solid. 

Ex.  1.  If  a  semicircle  is  conceived  as  revolved  around  its  diameter, 
what  is  the  path  through  which  it  moves  ?     See  Fig.  7. 

Ex.  2.  If  the  surface  OMNP,  Fig.  0,  is  conceived  as  revolved  around 
OP,  what  is  the  path  through  which  it  moves  ? 

Caution. — The  student  needs  to  be  careful  and  distinguish  between  the 
surface  traced  by  the  line  MN,  and  the  solid  traced  by  the  surface  OMNP. 

Ex.  3.  If  the  surface  represented  by  ABC  he  con- 
ceived as  revolved  about  its  side  CA,  what  kind  of 
a  solid  is  its  path  ? 

[Note.— As  has  been  said  before,  the  student  is  not 
necessarily  expected  to  luune  these  solids,  but  rather  to 
show,  in  his  own  language,  that  he  has  the  conception.] 

Ex.  4.  As  you  fill  a  vessel  with  water,  what  is  the  a 
solid  traced  by  the  surface  of  the  water  ? 
Ans.  The  same  as  the  space  within  the  yessel. 

Ex.  5.  If  a  circle  is  conceived  as  lying  horizontally,  and  then 
moved  directly  up,  what  will  be  the  solid  described,  t.  e.,  its  path  ? 
Do  not  confound  the  surface  described  with  the  solid.  What  de- 
scribes the  surface  ?     What  the  solid  ? 


Fig.  13. 


10  ELEMENTABY  GEOMETBY. 


EXTEXSIOX  AM)  FORM. 

28.  JExtension  means  a  stretching  or  reaching  out.  Hence,  a 
Point  has  no  extension.  It  has  only  position  (place).  A  Lme 
stretches  or  reaches  out,  but  only  in  length,  as  it  has  no  width. 
Hence,  a  hne  is  said  to  have  One  Dimension,  viz.,  length.  A  Surface 
extends  not  only  in  length,  but  also  in  breadth ;  and  hence  has 
Tico  Dimensions,  viz.,  length  and  breadth.  A  Solid  has  Tliree  Di- 
mensions, viz.,  length,  breadth,  and  thickness. 

III.— Suppose  we  think  of  a  point  as  capable  of  sti-etching  out  (extending) 
in  one  direction.  It  would  become  a  line.  Now  suppose  tlie  line  to  stretch  out 
(extend)  in  another  direction— to  widen.  It  would  become  a  surface.  Finaliy, 
suppose  the  surface  capable  of  thickening,  that  is,  extending  in  another  direc- 
tion.   It  would  become  a  sohd. 

29*  The  Limits  (extremities)  of  aline  are  points. 
The  Limits  (boundaries)  of  a  surface  are  lines. 
The  Limits  (boundaries)  of  a  solid  are  surfaces. 

30,  3Iaf/nitucle  (size)  is  the  result  of  extension.  Lines,  sur- 
faces, and  solids  are  the  geometrical  magnitudes.  A  point  is  not  a 
magnitude,  since  it  has  no  size.  The  magnitude  of  a  line  is  its 
length ;  of  a  surface,  its  area ;  of  a  solid,  its  volume. 

31,  Fiffure  or  Form  (shape)  is  the  result  of  position  of 
points.  The  form  of  a  line  (as  straight  or  curved)  depends  upon  the 
relative  position  of  the  points  in  the  line.  The  form  of  a  surface  (as 
plane  or  curved)  depends  upon  the  relative  position  of  the  points 
in  it  The  form  of  a  solid  depends  upon  the  relative  position  of  the 
points  in  its  surface.  Lines,  surfaces,  and  solids  are  the  geometrical 
figures.* 

III. — In  Fig.  14,  it  is  easy  to  conceive  the  form  of  the  lines  by  knowing  the 

position  of  points  in  the  lines.     By  taking  a 

quantity  of  common  pins  of  different  lengths, 

sticking  them  upright  in  a  board,  and  conceiv- 

1  /  \    ing  the  heads  to  represent  points  in  a  surface, 

V__^  (  I    we  can  readily  see  how  the  position  of  the  points 

in  a  surface  determine  its  form. 


Fig.  14. 


Ex.  1.  Suppose  a  line  to  begin  to  con- 


*  Line?,  surfaces,  and  solids  are  called  mao^nitadea  when  reference  is  had  to  their  extent^ 
and  fig:ure:^  w  hc-n  reference  is  had  to  their /or//i. 


SYNOPSIS.  11 

tract  in  length,  and  continue  the  operation  till  it  can  contract  no 
longer,  what  does  it  become  ?  That  is,  what  is  the  minor  limit  of  a 
line? 

Ex.  2.  If  a  surface  contracts  in  one  dimension,  as  width,  till  it 
reaches  its  limit,  what  does  it  become  ?  If  it  contracts  to  its  limit 
in  both  dimensions,  what  does  it  become  ? 

Ex.  3.  If  a  solid  contracts  to  its  limit  in  one  dimension,  what  does 
it  pass  into  ?    If  in  two  dimensions  ?     If  in  three  dimensions  ? 

Ex.  4.  What  kind  of  a  surface  is  that,  every  point  in  which  is 
equally  distant  from  a  given  point  ? 

S2.  Geometry  treats  of  magnitude  and  form  as  the  result  of 
extension  and  position. 

The  Geometrical  Concepts  are  points,  lines,  surfaces  (including 
plane  and  spherical  angles),  and  solids  (including  solid  angles). 

The  Object  of  the  science  is  the  measurement  and  comparison  of 
these  concepts. 

Plane  Geometry  treats  of  figures  all  of  whose  parts  are  confined  to  one  plane. 
Solid  Geometry,  called  also  Geometry  of  Space,  and  Geometry  of  Three  Dimensions, 
treats  of  figures  whose  parts  lie  in  different  planes.  The  division  of  Part  II. 
into  two  chapters  is  founded  upon  this  distinction.  In  the  Higher  or  General 
Geometiy  these  divisions  are  marked  by  the  terms  "  Of  Loci  in  a  Plane,''  and 
"  Of  Loci  in  Spaced 


o 

JO 

< 


A 


SYNOPSIS. 

What. — How  designated. — 111. 


Point \  Dimensions  of. 

Limit  of  Line. — Surface. — Solid. 


Line 


r  "What. 
How  designated. 
Dimensions  of. 
Limit  of  Surface, 

(  Straight.— What.— 7^?. 
Kinds  \  Curved.— What.— I/^. 

(  Broken  (?). 

r  What. 
Dimensions  of. 
Limit  of  Solid. 
T^;„ria  J  Plane.— What— 72?. 
J^mas  ^  Curved.— What.— 7?Z. 

iWhat. — Size  depends  on  what Adjacent. 
(  Right.— What.— //?. 
Kinds  \  ni.i:^„^  j  Acute.-What— //?. 
I  Oblique  J^  obtuse.-What.-/??. 

Solid What— 7ZZ.— Examples. 

Macmitude.- What— Result  of  what 


Surface 


r  rr      f     p  S  Macmitude.- What— Result  ot  what 

I  >  ^'^^^^  ^^  \  Figure  or  form.— AVhat.— Result  of  what 
\  Concepts. — Wh 
t  Object— What 


Geometry...  Concepts.-What 


PART  L 


A  FEW  OF  THE  MORE  IMPORTANT  FACTS  OF  THE 

SCIENCE. 


Ai 

^B 

p, 

-HQ 

p, 

-H 

li 

SECTION  L 

ABOUT  STRAIGHT  LINES. 

SS*  JProh, — To  measure  a  straight  line  with  the  dividers  and 
scale. 

Solution.— Let  AB,  Fig.  15,  be  the  line  to  be  measured.    Take  the  dividers, 

Fig.  2  (frontispiece),  and  placing 

the  sharp  point  A  firmly  upon 

the  end  A  of  the  line  AB,  open 

the  dividers  till  the  other  point 

B  (the  pencil  point)  just  reaches 

the  other  end  of  the  line  B. 

Then  letting  the  dividers  re- 

)K    main  open  just    this  amount, 

^^^-  ^^-  place  the  point  A  on  the  lower 

end  of  the  left  hand  scale,  as  at  o,  Fig.  1,  and  notice  where  the  point  B  reaches. 

In  this  case  it  reaches  3  spaces  beyond  the  figure  1.    Now,  as  this  scale  is 

iiXiihes  and  tentJis  of  inches,*  the  Ime  AB  is  1.3  inches  long. 

Ex.  1.  What  is  the  length  of  CD  ?  A7is.  .15  of  a  foot 

Ex.  2.  What  is  the  length  of  EF  ?  Ans.  .75  of  an  inch. 

Ex.  3.  What  is  the  length  of  CH  ?  Ans.  H  inches. 

Ex.  4.  What  is  the  length  of  IK  ?  Ans.  .18  of  a  foot. 

Ex.  5.  Draw  a  line  3  inches  long. 
Ex.  6.  Draw  a  line  2.15  inches  long. 
Ex.  7.  Draw  a  line  1.25  inches  long. 
Ex.  8.  Draw  a  line  .85  of  an  inch  long. 

*  The  next,  scale  to  the  right  is  divided  into  lOths  and  lOOths  of  a  foot.    Thus  frouxp  to  IC 
1  tentl*  of  a  foot.  ai.d  the  pmaller  divisions  are  hundredths. 


ABOUT  STRAIGHT  LINES.  13 

[Note. — Suppose  a  fine  elastic  cord  were  attached  by  each  of  its  ends  to  the 
points  A  and  B  of  the  dividers  ;  when  they  were  opened  so  as  to  reach  from 
C  to  D,  Fig.  15,  the  cord  would  represent  the  line  CD,  Now  applying  the  di- 
viders to  the  scale  is  the  same  as  laying  this  cord  on  the  scale.  Without  the 
cord,  we  can  imagine  the  distance  between  the  points  of  the  dividers  to  be  a  line 
of  the  same  length  as  CD.] 

Ex.  9.  Find  in  the  same  way  as  above  the  length  and  width  of  this 
page.  Also  the  distance  from  one  corner  (angle)  to  the  opposite  one 
(the  diagonal). 


34:,  IProb, — To  find  tlie  sum  of  tivo  lines. 

Solution. — To  find  the  sum  of  AB  and  CD,  1*  first  draw  the  indefinite  line 
Ejx.    With  the  dividers  I  obtain  the  length  of  AB,  by  placing  one  point  on  A 

and  extending  the  other  to  B,     A' 'B 

This  length  I  now  lay  ofi"  on  the 
indefinite  line  E^,  by  putting  one 
point  of  the  dividers  at  E  and     E 


C>- 


F  G 

with  the  other  marking  the  point  Fig.  ig. 

F.     EF  is  thus  made  equal  to  AB. 

In  the  same  manner  taking  the  length  of  CD  with  the  dividers,  I  lay  it  off  from 

F  on  the  line  F.r.    Thus  I  obtain  EG=EF  +  FC=AB  +  CD.     Hence,  the  sum  of 

AB  and  CD  is  EG. 

[Note.— The  student  may  measure  EG  by  (55)  and  find  the  sum  of  AB  and 
CD  in  inches  or  feet;  but  it  is  most  important  that  he  be  able  to  look  upon  EG 
as  the  sum  itself.] 

Ex.  1.  Find  the  sum  of  AB  and  EF,  Fig.  15. 
Ex.  2.  Find  the  sum  of  EF,  CD,  and  GH,  Fig.  15. 

Ex.  3.  Make  a  line  twice  as  long  as  CD,  Fig.  16.  Three  times  as 
long. 


'  35.  JProb, — To  find  the  difference  of  two  lines. 

Solution.— To  find  the  difference  of  AB  and  CD,  I  take  the  length  of  the 

less  line  AB  with-the  dividers  ;  and  placing 

one  point  of  the  dividers  at  one  extremity 

of  CD,  as  C,  make  Ce  =  AB.    Then  is  eD     c« • «D 

the  difference  of  AB  and  CD,  since  eD  =  Fi«.  17. 

CD  -  C(3  =  CD  -  AB. 

Ex.  1.  Find  the  difference  of  IK  and  EF,  Fig.  15. 
Ex.  2.   Find  the  difference  of  GH  and  CD,  Fig.  15. 

*  Theee  elementary  solutions  are  sometimes  put  in  the  singular,  as  the  more  simple  style. 


14  ELEMENTARY  GEOMETEY. 

Ex.  3.  Find  how  much  longer  IK,  Fig.  15,  is  than  the  sum  of  EF, 
Fig.  15,  and  CD,  Fig.  16. 

Ex.  4.  Find  tlie  difference  of  the  sum  of  AB  and  CH,  and  the 
sum  of  CD  and  EF,  Fig.  15. 


36.  Prob. — To  compare  the  lengths  of  two  lines  ;  that  is,  to  find 
their  ratio  (approximately*). 

Solution. — To  compare  the  lengths  of  AB  and  CD,  I  lay  off  AB,  the  shorter, 

upon  CD,  as  Ca.    (If  AB  could  be 

C" — ^ — y- — y^D     applied  two  or  more  times  to  CD, 

I  should  apply  it  as  many  times  as 

^  <p  CD  would  contain  it)    Now  I  apply 

*"  J     .>  /S  the  remainder  of  CD,  viz.,  aD,  to  AB, 

^^^-  ^^-  as  many  times  as  AB  will  contain  it, 

which  is  once  with  the  remainder  5B.  This  remainder  I  now  appl}"  to  <tD,  and 
find  it  contained  once  with  a  remainder  cD.  Again,  I  apply  this  last  remainder 
to  6B,  and  find  it  contained  twice  with  a  remainder  dB.  This  last  remainder  I 
now  apply  to  cD,  and  find  it  contained  3  times,  without  any  remainder.  This 
last  measure,  cZB,  is  a  common  measure  of  the  two  lines.  Calling  dB  1, 1  now 
observe  that 

fZB  =  1 ; 

cD  =MB=  3; 

hd    =2cD=6; 

ac    =bS  r=  hd  +  dB  =  7; 

«D  =  ac  +  cD  =  10; 

AB  =  A&  4-  6B  =  aD  +  ac  =  17; 

CD  =  Ca  +  aD  =  AB  +  «D  =  27. 

Hence  the  lines  AB  and  CD  are  to  each  other  as  the  numbei*s  17  and  27 ;  AB 
is  i^  of  CD;  or,  expressed  in  the  form  of  a  proportion,  AB  :  CD  : :  17  :  27. 

[Note. — This  process  will  be  seen  to  be  the  same  as  that  developed  in  Arith- 
metic and  Algebra  for  finding  the  greatest  or  highest  Common  Measure  of  two 
numbers,  and  should  be  studied  in  connection  with  a  review  of  those  processes. 
See  Complete  Aritidietic  {115),  and  Complete  School  Algebra  (J.57).] 

Ex.  1.    Find,  as  above,  the  approximate  ratio  of  AB  to  CD,  Fig.  15. 

Ratio,  n:\^. 
Ex.  2.  Find,  as  above,  the  approximate  ratio  of  CD  and  IK,  Fig.  15. 

Ratio,  5 : 6. 


*  This  method  does  not  get  the  exact  ratio,  because  of  the  imperfection  of  measurement,  and 
also  because  lines  are  sometimes  incommensurable,  as  will  appear  hereafter. 


ABOUT   STRAIGHT  LINES. 


15 


Ex.  3.  Find,  as  above,  the  approximate  ratio  of  EF  to  CH,  Fifj.  15. 

Ratio,  1 : 2. 

Ex.  4.  Find,  as  above,  the  approximate  ratio  of  EF  to  CD,  Fig.  15. 

Ratio,  5 :  12. 


57.    To  Intersect  is  to   cross;   and  a  crossing  is  called  an 
Inter  sectiofi. 

38,   To  JBisect  anything  is  to  divide  it  into  two  equal  parts. 
5.9.  Prob»—To  hised  a  given  line. 


Fig.  19. 


Solution. — To  bisect  the  line  AB,  I  take  the  dividers;  and  opening  them 
so  that  the  line  between  their  points  is  more  than  ^^^ 

half  as  long  as  AB,  I  place  the  sharp  point  A  on 
the  point  A,  and  holding  it  firmly  there,  make  a 
little  mark  with  the  pencil  point  B,  as  nearly  as  I 
can  guess,  opposite  the  middle  of  the  line.  Then, 
being  careful  to  keep  the  dividers  open  just  the 
same,  I  place  the  sharp  point  on  B,  and  make  a 
mark  intersecting  the  first  one,  as  at  m.  Now, 
doing  just  the  same  on  the  other  side  of  the  line, 
I  make  two  marks  intersecting  each  other,  as  at  n. 
Finally,  I  draw  a  line  from  m  to  n,  and  where  this 
line  crosses  AB  is  its  middle  point;  that  is,  AO  is  equal  to  OB 
so  we  do  not  propose  to  tell  now 
it. 


[Why  this  is 
The  student  needs  only  to  learn  how  to  do 
He  should  measure  AO  and  OB,  and  thus  test  the  accuracy  of  his  work.] 


Ex.  1.  Is  it  necessary  that  the  dividers  be  opened  just  as  wide 
when  the  marks  are  made  through  n,  as  when  they  are  made 
through  m?     Try  it. 

,  Ex.  2.  Suppose  you  make  the  marks  through  m  as  directed,  but, 
in  making  those  through  n^  you  have  the 
dividers  ivider  open  when  you  put  the  point 
on  A  than  when  you  put  it  on  B;  will  the 
line  joining  m  and  n  then  cross  AB  in  the 
middle  ?  If  not,  on  which  side  of  the  mid- 
dle will  0  be  ?     Try  it. 

Ex.  3.   Can  you  bisect  a  line  by  making     ^ 
the  marks  all  on  one  side  of  it  ?  If  so,  do  it. 


o 

Fig.  20. 


16  ELEMENTARY  GEOMETRY. 

40.  Axiom* — A  straight  line  is  the  shortest  path  between  two 
points. 

III.— If  a  cord  is  stretched  across  the  table,  it  marks  a  straight  line.  In  this 
way  the  carpenter  marks  a  straight  line.  Having  rubbed  a  cord,  called  a  chalk- 
line,  with  chalk,  he  stretches  it  tightly  from  one  point  to  another  on  the  surfaces 
upon  which  he  wishes  to  mark  the  line,  and  then  raising  the  middle  of  the 
cord,  lets  it  snap  upon  the  surface.  So  the  gardener  makes  the  edges  of  his 
paths  straight  by  stretching  a  cord  along  them.  These  operations  depend  upon 
the  principle  that  when  the  line  between  the  points  is  the  shortest  possible,  it 
is  straight 

41.  Axiom, — Tico  points  in  a  straight  line  determine  its 
position. 

III.— If  the  farmer  wants  a  straight  fence  built,  he  sets  two  stakes  to  mark 
its  ends.  From  these  its  enth-e  course  becomes  known.  This  is  the  principle 
upon  which  aligning  (or  sighting)  depends.  Having  given  two  points  in  the 
required  hne,  by  looking  in  the  direction  of  one  from  the  other,  we  look  along  a 
straight  line,  and  are  thus  able  to  locate  other  points  in  the  line.     If  the  points 

A  and  B  are  marked,  by 

^  putting  the  eye  at  A  and 

^^K D ' — B <^ ~ — fe      looking    steadily  towards 

*C  B,  we  can  tell  whether  D 

^^^-  ^^"  and   E  are   in    the   same 

straight  line  with  A  and  B,  or  not  So  we  can  observe  that  C  and  C"  are  not 
in  the  line ;  but  that  C  is.  This  process  of  discovering  other  points  in  a  line 
with  two  given  points  is  called  aligning,  or  sighting.  In  this  way  a  row  of 
trees  is  made  straight,  or  a  line  of  stakes  set  It  is  the  principle  upon  which 
the  surveyor  runs  his  lines,  and  the  hunter  aims  his  gun.  In  the  latter  case, 
the  two  sights  are  the  given  points,  and  the  mark,  or  game,  is  a  third  point, 
which  the  maiksman  wishes  to  have  in  the  same  sti'aight  line  as  the  sights. 


42,  Axiom, — Between  the  same  tivo  points  there  is  one  straight 
line,  and  only  one. 

III.— Let  any  two  lettei-s  on  this  page  represent  the  situation  of  two  points ; 
we  readily  see  that  there  is  one,  and  only  one,  straight  path  between  them. 
Again,  let  a  corner  of  the  desk  represent  one  point  and  a  corner  of  the  ceiling 
of  the  room  represent  another  point  ;  we  perceive  at  once  that,  if  a  point  is 
conceived  to  pass  in  a  straight  line  from  one  to  the  other,  it  will  always  trace 


•  An  axiom  may  be  illustrated,  but  it  needs  no  demonstration.  We  may  explain  the  terras 
used  and  elaborate  the  condensed  etatement ;  but  if,  when  its  meaning  is  clearly  understood, 
any  one  does  not  grant  the  truth  of  its  statement,  he  has  not  a  sound  mind,  and  we  cannot 
reason  with  him. 


ABOUT  STRAIGHT  LINES.  17 

the  same  path.  In  short,  as  soon  as  two  j^oints  are  mentioned,  we  think  of  the 
distance  between  them  as  a  single  straight  line, — for  example,  the  centre  of  the 
earth  and  the  centre  of  the  sun. 

Once  more,  conceive  A  and  B,  Fig.  21,  to  be  two  points  in  the  path  of  a  point 
moving  from  A  in  the  direction  of  B.  Now  all  the  points  in  the  same  direction 
from  A  as  B  is,  are  in  this  path ;  and  any  point  out  of  this  line,  as  C  or  C",  is 
in  a  different  direction  from  A. 

In  this  manner  we  draw  a  straight  line  on  paper  by  laying  the  straight  edge 
of  a  ruler  on  two  points  through  which  we  wish  the  line  to  pass,  and  passing  a 
pen  or  pencil  along  this  edge. 

Cor. — Tiuo  straigM  lines  cayi  intersect  in  hut  one  point ;  for,  if 
they  had  tiuo  points  common,  they  would  coincide  and  not  intersect. 

Ex.  1.  A  railroad  is  to  be  run  from  the  town  A  to  town  B.  If  it  is 
made  straight,  through  what  points  will  it  pass  ?  Can  it  pass  through 
any  points  not  in  the  same  direction  from  A  as  B  is  ? 

Ex.  2.  If  I  live  on  the  south  side  of  a  straight  railroad,  and  my 
friend  on  the  north  side,  but  five  miles  farther  east,  and  two  miles 
farther  north,  and  the  road  from  my  house  to  his  is  straight,  how 
many  times  does  it  cross  the  railroad  ? 

Ex.  3.  Can  you  always  draw  a  straight  line  which  shall  cut  a 
curve  (whatever  curve  it  may  be)  in  two  points  ?     Try  it. 

Ex.  4.   Detroit  is  directly  east  of  where  I  live.    How  could  I  drive 
my  horse  there  and  never  turn  his  head  to  the  east  ?  Would  he  have 
to  travel  in  straight  lines  or  in  a  curve?    If  I  drive  him  on  a  curve,, 
how  can  I  manage  it  so  that  his  head  will  be 
east  for  but  an  instant?    If  his  head  is  all  (x^ 

the  time  east,  what  is  the  line  in  which  I  ..-''"'       \ 

drive  him  ?  /'''  \ 

P^^ ^Q. 

SuG.— The  figure  will  suggest  how  the  first  may  ^m   22 

be  accomplished. 


45.  A  JPerpendicular  to  a  given  line  is  a  line  wliicli  makes 
a  right  angle  {26)  with  the  given  line.  The  latter  is  also  perpen- 
dicular to  the  former.  Oblique  Lines  are  such  as  are  not  perpen- 
dicular to  each  other,  and  which  meet  if  sufficiently  extended. 

III.— In  Fig.  11,  BA  is  perpendicular  to  DC  ;  so  also  AC  is  perpendicular  to 
BA.  In  Fig.  10,  KG  and  Kl  are  perpendicular  to  each  other.  The  other  lines 
in  Fig.  10  are  oblique  to  each  other. 

2 


18 


ELEMENTARY  GEOMETRY. 


6X 


A 
Fig.  23. 


44.  I^rob* — To  erect  a  2^er2)endicular  to  a  given  li7w  at  a  given 
poi7it  in  the  line. 

Solution. — Suppose  I  want  to  erect  a  perpendicular  to  the  line  XY,  at  the 

point  A.  With  the  dividers  I  measure 
off  a  distance  AB  on  one  side  of  the  point 
A,  and  an  equal  distance  AC  on  the  other 
side.  Then  opening  the  diridei-s  a  little 
•wider,  I  put  the  sharp  point  on  B  and 
^  V     make  a  mark  with  the  pencil  point,  as 

at  0,  about  where  I  think  the  perpen- 
dicular will  go.  Then,  keeping  the  dividers  open  just  tlie  same,  I  put  the  shai-p 
point  on  C,  and  make  a  mark  intersecting  the  former  one  at  0.  Now,  drawing 
a  line  through  O  and  A,  it  is  the  perpendicular  sought. 

Ex.  1.  Suppose  I  make  a  mistake  and  close  up  the  dividers  a 
little  after  making  the  first  mark  through  0,  and  then  make  the  sec- 
ond mark  ;  which  way  will  the  line  lean  ?  Will  it  be  a  pei-pendicu- 
lar  or  an  oblique  line  in  this  case  ?  What  kind  of  an  angle  would 
OAY  be  ?  What  OAX  ?  What  kind  of  angles  are  these  when  OA  is  a 
perpendicular  ? 

Ex.  2.  Suppose  I  should  mistake  a  point  nearer  to  A  than  B  was 
taken,  and  use  it  as  I  did  C,  having  the  dividers  open  just  alike  when 
I  made  the  two  marks  through  0 ;  which  way  would  the  line  lean 
(incline)  ?     (Same  questions  as  in  the  last.) 


V     B 


M 


Fig.  24. 


N 


III. — A  carpenter  wishes  to  get  the 
piece  of  timber  AF  at  right  angles  *to 
MN,  into  which  it  is  mortised  at  A.  So  he 
measures  off  AB  and  AC,  equal  distances 
from  A ;  and  taking  two  poles  of  equal 
length  (say  10  feet  long),  has  the  end  of  one 
held  steadily  at  B  and  the  end  of  the  other 
at  C,  and  moves  (racks,  as  he  calls  it)  the 
end  F  to  the  right  or  left  until  the  ends  E 
and  D  of  the  poles  are  exactly  opposite,  as 
in  the  figure.  AF  is  then  perpendicular  to 
MN. 


45.  Proh* — From  a  point  without  a  given  line,to  draw  a  perpen- 
dicular to  the  line. 


ABOUT   CIRCLES. 


19 


Solution.— I  wish  to  draw  a  perpendicular  from  O  to  the  line  XY.     I  first 
open  the  dividers  wide  enough,  so  that  when  I  q 

place  the  sharp  point  on  0  the  pencil  will  mark 
the  line  XY  in  two  points,  as  B  and  C,  when  it 
swings  around.  Marking  these  two  points,  I 
put  the  sharp  point  first  on  B  and  afterward  on 
C,  keeping  them  open  just  alike  in  both  cases,  -y^ — ^ 
and  make  the  two  marks  intersecting  at  D. 
Placing  the  straight  edge  of  the  ruler  on  the 
points  O  and  D,  I  draw  the  line  OA  along  its 
edge.     OA  is  the  perpendicular  required. 


A 


0/ 


^^ 


Fig.  25. 


Ex.  1.  Let  fall  a  perpendicular  from  a  point,  as  0,  upon  a  straight 
line,  as  XY,  without  making  any  marks  on  the  opposite  side  of  XY 
from  o. 

Ex.  2.  A  mason  wishes  to  build  a 
wall  from  o,in  the  wall  A B,"  straight 
across"  (perpendicular)  to  the  wall 
CD,  which  is  8  feet  from  AB.  He  has 
only  his  10-foot  pole,  which  is  subdi- 
vided into  feet  and  inches,  with  which 
to  find  the  point  in  the  opposite  wall  at  which  the  cross  wall  must 
join.    How  shall  he  find  it  ? 


SECTION  IL 

ABOUT  CmCLES. 


46*.  A  Circle  is  a  plane  surface  bounded  by  a  curved  line  every 
point  in  which  is  equally  distant  from  a  point  within. 

47.  The  Circiitnference  of  a  Circle  is  the  curved  line  every 
point  in  which  is  equally  distant  from  a  point  within. 

4:8.  The  Centre  of  a   Circle  is  the  point  within,  which   is 
equally  distant  from  every  point  in  the  circumference. 

4i9.  An  Arc  is  a  part  of  a  circumference. 

50.  A  Hadilis  is  a  line  drawn  from  the  centre  to  any  point 
in  the  circumference  of  a  Circle. 

51,  A  I>ianiefer  of  a  Circle  is  a  line  passing  through  the 
centre  and  terminating  in  the  circumference. 


20 


ELEMENTARY  GEOMETRY. 


Fig.  27. 


Ii.L. — A  circle  may  be  conceived  as  the  path  of  a 
line,  like  OB,  Fig.  27,  one  end  of  which,  O,  remains 
at  the  same  point,  while  the  other  end,  B,  moves 
around  it  in  the  plane  (say  of  the  paper).  OB  is  the 
Badius,  and  the  path  described  by  the  point  B  is  the 
Circumference.  AB  is  a  diameter.  In  Fig.  28,  the 
curved  line  ABCDA  (going  clear  around)  is  the  Qir- 
cumference,  0  is  the  Centre,  and  the  space  within  the 
circumference  is  the  Circle.  Kiaj  part  of  a  circum- 
ference as  AB,  or  any  of  the  curved  lines  BB,  Fig.  27, 
is  an  arc.  So  also  AM  and  EF,  Fig.  29,  are  arcs.  EF 
is  an  arc  drawn  from  O'  as  a  centre,  with  the  radius 
O'B. 

52.  A  Chord  is  a  straight  line  joining 
any  two  points  in  a  circumference,  but  not 
passing  through  the  centre,  as  BC  or  ad. 
Fig.  28.  The  portion  of  the  circle  included 
between  the  chord  and  its  arc,  as  kmD,  is  a 
Segment. 


53,  A  Tangent  to  a  circle  is  a  straight 
line   which  touches  the    circumference,   but 
Fig.  29.  does  uot  iutcrsect  it,  how  far  soever  the  line 

be  produced. 

54:,  A  Secant  is  a  straight  line  which  intersects  the  circumfer- 
ence in  two  points. 


Ex.  1.  Suppose  DC,  Fig.  11,  to  represent  a  small  wooden  rod,  and 
BA  a  wire  stuck  into  it  at  right  angles.  Now  if  you  take  the  end  C 
of  the  rod  in  your  fingers  and  place  the  end  D  on  the  table  so  that 
the  rod  shall  stand  upright,  and  then  revolve  the  rod  once  around 
like  a  shaft,  what  will  the  wire  describe  ?  What  the  end  B  ?  What 
any  point  in  BA  ?  If  you  only  revolve  the  rod  a  little  way,  what  will 
the  point  B  describe  ?     What  does  BA  represent  ? 

Ex.  2.  If  you  take  a  string,  OP,  and  hold  one  end  at  a  particular 
point,  O,  on  your  slate  or  blackboard,  while 
with  the  other  hand  you  hold  the  other 
end,  p,  of  the  string  upon  the  end  of  a 
pencil  or  crayon,  and  then  move  the  end 
P  around  0,  making  a  mark  as  it  goes, 
what  will  the  mark  made  represent  when 
the    pencil    or    crayon    has    gone    clear 


^^x 


Pro.  so. 


ABOUT   CIECLES.  21 

around  ?    What  will  the   string  represent  ?     What  is  the  surface 
passed  over  by  the  string  ? 

Ex.  3.  If  you  take  the  dividers,  Fig.  1,  and  open  them  (say  2 
inches),  and  then  place  the  sharp  point.  A,  firmly  on  the  paper  while 
you  turn  them  around,  making  the  pencil  point,  B,  mark  the  paper 
as  it  goes,  what  kind  of  a  line  will  be  described  ?  What  is  the  line 
joining  the  points  of  the  dividers  ?*  What  line  describes  the  cir- 
cle ?  If  the  dividers  only  turn  a  little  way,  what  is  the  line 
described  ? 

Ex.  4.  If  a  boy  skating  on  the  ice  makes  a  curve  which  bends 
everywhere  just  alike,  Avhat  kind  of  a  path  will  he  make  ?  Does 
the  boy  describe  a  circle  ?  How  might  you  conceive  the  circle  in- 
closed by  his  path,  as  described  ?  Is  a  circle  described  by  a  point  or 
by  a  line  ? 

[Note. — The  word  "circle"  is  used  in  common  language  as  equivalent  to 
"  ciicumference."  It  is  also  thus  used  in  General  Geometry.  But,  however  the 
words  may  be  used,  the  pupil  should  be  taught  to  mark  the  distinction  between 
the  plane  surface  inclosed  and  the  bounding  line.] 

Ex.  5.  In  how  many  points  can  a  straight  line  intersect  a  circum- 
ference? In  how  many  points  can  one  circumference  intersect 
another  ? 

Ex.  6.  There  is  a  piece  of  ground  in  the  form  of  a  circle,  the 
radius  of  which  is  100  rods,  by  which  run  two  roads;  one  road 
runs  within  80  rods  of  the  centre,  and  the  other  within  100  rods. 
How  do  the  roads  lie  with  reference  to  the  ground  ? 

Ex.  7.  When  you  unwind  a  thread  by  drawing  it  off  a  spool  in 
the  ordinary  way,  what  geometrical  line  does  the  unwound  thread 
represent  ? 

"Ex.  8.  In  a  circle  whose  diameter  is  50  feet,  there  are  drawn  two 
chords,  one  is  20  feet  long,  and  the  other  30  feet.  Which  is  nearer 
the  centre  ? 

Ex.  9.  There  are  two  circles  whose  radii  are  respectively  12  and 
18  feet.  The  distance  from  the  centre  of  one  to  the  centre  of  the 
other  is  25  feet.  Do  the  circumferences  intersect  ?  Would  they  in- 
tersect if  the  centres  were  3  feet  apart  ?  How  would  they  lie  in  ref- 
erence to  each  other  in  the  latter  case  ?  How  if  their  centres  were 
30  feet  apart  ?     How  if  they  were  35  feet  apart  ? 

*  The  imagination  may  be  aided  by  supposing  a  tine  elastic  cord  stretched  between  the 
points  of  the  dividers  and  carried  by  them. 


22 


ELEMENTARY  GEOMETRY. 


Ex.  10.  What  kind  of  a  line  is  represented  by  water  flying  from  a 
swiftly-revolving  grindstone  ? 

Ex.  11.  If  you  draw  two  chords  in  the  same  circle,  one  of  which 
is  twice  as  long  as  the  other,  will  the  arc  cut  off  by  the  longer  chord 
be  twice  as  long  as  the  arc  cut  off  by  the  shorter  ?  Will  it  be  more 
than  twice  as  long,  or  less  ? 


55,  TJieorem* — The  cliord  of  a  sixth  part  of  the  circumference 
of  a  circle  is  just  equal  to  the  radius  of  the  same  circle. 

III. — If  I  draw  a  circle,  and  then,  being  careful  not  to  open  or  close  the  di- 
viders, place  the  shaip  point  on  tlie  circumference 
at  some  point,  as  A,  and  mark  the  cu'cumferencc  at 
another  point,  as  B,  with  the  pencil  point,  and  then 
move  the  sharp  point  to  B  and  mark  again,  as  C,  1 
find  that  when  I  have  measm'ed  oflF  six  such  chords, 
each  equal  to  the  radius,  I  return  exactly  to  A,  the 
point  of  starting. 

Moreover,   if  I  draw  the  chords  AB,  BC,  etc.,  I 

have  a  regular  figure  with  six  equal  sides.    A  figure 

^^^-  31-  with  six  sides  is  called  a  hexagon.    This  hexagon  is 

called  regular,  because  its  sides  are  equal  each  to  each,  and  its  angles  are  also 

mutually  equal. 

Again,  if  I  unite  the  alternate  angles  of  the  regular  hexagon,  as  FB,  BD,  and 
DF,  I  have  a  regular  triangle,  called  an  equilateral  triangle. 

56,  Inscribed  Figures  are  figures  drawn  in  a  circle,  and 
having  the  vertices  of  all  their  angles  in  the  circumference,  as  the 
hexagon  and  triangle  in  the  last  illustration.  When  the  figure  is 
without,  and  all  its  sides  touch  but  do  not  cut  the  circumference,  it 
is  circumscribed  about  the  circle. 

Ex.  1.  Draw  a  regular  hexagon  whose  side  is  two  inches. 

Ex.  2.  Inscribe  an  equilateral  triangle  in  a  circle  whose  radius  is 
one  inch. 


57,  JProh, — To  find  the  centre  of  a  circle  ichen  the  circumference 
is  draivn  (or,  as  we  usually  say,  known). 

Solution. — The  circumference  of  my  circle  is  drawn,  but  the  centre  is  not 


ABOUT  CIRCLES. 


23 


marked.  So  I  want  to  find  the  centre.  I  draw 
any  two  chords,  as  AB  and  CD  (the  nearer  they  are 
at  right  angles  to  each  other  the  better  for  accu- 
racy). I  then  bisect  each  chord  with  a  perpen- 
dicular, as  AB  with  the  perpendicular  MN,  and 
CD  with  RS  {39).  The  intersection  of  these  two 
perpendiculars,  as  0,  is  the  centre  of  the  circle. 
[The  pupil  must  do  everything  with  his  pencil, 
ruler,  and  dividers,  just  as  he  says.  He  must  not 
be  of  those  who  "  say  and  do  not."    He  must  do  the  Fig.  32. 

things  told,  *'  over  and  over,"  till  he  can  do  them  neatly  and  easily.] 


S8,  JProb, — To  pass  a  circumference  through  three  given  pomts. 

Solution.— I  wish  to  pass  a  circumference  through  the  three  given  points 
A,  B,  and  C.  [The  pupil  should  first  designate  three 
points  by  dots  on  his  paper,  slate,  or  board,  and  then 
proceed  according  to  the  solution.]  In  order  to  do  this, 
I  join  A  and  B  with  a  line,  and  also  B  and  C.  I  now 
bisect  these  lines  with  the  perpendiculars  MN  and  RS, 
as  in  the  last  problem.  The  intersection  of  these  per- 
pendiculars, 0,  is  the  centre  of  the  required  circle. 
Now  setting  the  sharp  point  of  the  dividers  upon  0  and  ^ 

opening  them  till  the  pencil  point  just  reaches  A  (B  or  "N^ 

C  will  answer  as  well),  I  draw  the  circumference  with  Fig.  33. 

0  as  its  centre  and  the  radius  OA,  and  find  that  it  passes  through  the  three 
given  points  A,  B,  and  C. 

Ex.  1.  To  pass  a  circumference  tlirougli  the  three  vertices  of  a 
triangle,  i,  e.,  to  circumscribe  a  circumference  about  a  triangle,  as 
this  operation  is  technically  called. 

SuG.— This  is  just  like  the  last,  A,  B,  and  C  being  the  vertices  of  the  triangle. 
The  four  figures  in  the  margin 
represent  the  successive  steps  in 
the  solution.  First  draw  the  given 
triangle.  Then  take  the  first  step 
in  the  solution,  then  the  second, 
etc. 

Ex.  2.  Given  the  centre  of 
a  circle  and  a  point  in  the 
circumference,  to  draw  the 
circle. 

SuG. — Make  a  dot  on  the  board 
to  indicate  the  centre,  and  an- 
other dot  to  indicate  the  point 
in  the  circumference  to  be  found. 
This  is  what  is  given.    You  are 


Fi?.  34. 


24  ELEMENTAKY   GEOMETRY. 

then  to  draw  the  circumference,  which  shall  pass  through  the  latter  point,  and 
have  the  former  for  its  centre. 

Ex.  3.  Draw  an  arc  of  a  circle,  and  rub  out  the  mark,  if  you  make 
any,  at  the  centre,  so  that  you  cannot  see  where  the  centre  is.  Then 
find  the  centre,  and  complete  the  circumference  according  to  these 
problems. 

SuG. — Mark  three  points  in  the  given  are,  and  tnen  the  example  is  just  like 
the  last  [Do  not  fail  to  do  it,  "  over  and  over,"  till  you  can  do  it  quickly  and 
neatly.     These  exercises  require  much  care  in  order  to  get  good  figures.] 


S9»  Tlieoreni, — Tlie  circumference  of  a  circle  is  ahciit  3.1416 
times  its  diameter.  The  Greek  letter  n  (called  p)  is  used  to  repre- 
sent this  niimher  ;  and  hence  the  circumference  is  said  to  be  n  times 
the  diameter, 

III. — The  pupil  can  illustrate  this  fact  by  taking  any  wheel  which  is  a  true 
circle,  and  measuring  the  diameter  with  a  narrow  band  of  paper  (something 
that  will  not  stretch),  and  then  wrapping  this  measure  about  the  circumference. 
He  will  find  that  it  takes  a  little  more  than  three  diameters  to  go  around.  Of 
course  he  cannot  tell  exactly  how  much  more.  In  fact,  nobody  knows  exactly. 
But  the  number  given  above  is  near  enough  for  most  purposes.  For  many  pur- 
poses 3f  is  sufllciently  accurate. 

By  drawing  a  circle  veiy  carefully,  say  1  inch  in 
diameter,  as  in  the  margin,  and  dividing  the  diameter 
into  lOths  inches,  a  nice  pair  of  dividers  can  be 
opened  one  10th  inch  and  made  to  step  around  the 
circumference.  If  it  is  all  done  with  nicety,  it  wUl  be 
found  to  be  a  little  over  31  steps  around,  when  it  is 
10  across. 

Ex.  1.  The  distance  across  a  wagon-wheel 
Fig.  a3.  (the  diameter)  is  4  feet,  how  long  a  bar  of  iron 

will  it  take  to  make  the  tire  ? 

Ex.  2.  Suppose  the  crown  of  your  hat  is  a  circular  cylinder  7 
inches  in  diameter,  how  much  ribbon  Avill  it  take  for  a  band,  allow- 
ing \  of  a  yard  for  the  knot  ? 

Ex.  3.  How  many  times  will  the  driving-wheel  of  an  engine,  which 
is  6  feet  in  diameter,  revolve  in  going  from  Detroit  to  Chicago,  a 
distance  of  288  miles,  allowing  nothing  for  slipping  ? 

Ex.  4.  A  boy's  hoop  revolved  200  times  in  going  around  a  city- 
square,  a  distance  of  140  rods.     What  was  the  diameter  of  his  boon  ? 


ABOUT  ANGLES. 


25 


Ex.  5.  What  is  the  radius  of  a  circle  whose  semi-circumference  is 
;r  ?     In  -a  circle  whose  radius  is  1,  what  part  of  the  circumference 

7t  7t 

does  -X  represent  ?    What  part  —  ?    What  part  does  %7t  represent  ? 


SECTION  IIL 


ABOUT  ANGLES. 
60.  I^roh, — To  sJioiv  koto  angles  are  generated  and  measured. 

III. — An  angle  is  generated  by  a  line  revolving  about  one  of  its  extremities. 
Thus,  suppose  OB  to  have  started  from  coincidence  with  OA,  and,  0  remaining 
fixed,  the  line  to  have  revolved  to  the  position  OB, 
the  angle  BOA  would  have  been  generated.  When 
the  revolving  line  has  passed  one-quarter  the  way 
around,  as  to  DO,  it  has  generated  a  right  angle ; 
when  one-half  way  around,  as  to  FO,  two  right 
angles  ;  when  entirely  around,  four  right  angles.      ^ 

Now,  if  any  circle  be  described  from  0  as  a  cen- 
tre, the  arc  included  by  the  sides  of  any  angle  having 
its  vertex  at  0,  is  (he  same  part  of  a  quarter  of  this 
circumference  as  the  angle  is  of  a  right  angle.  Hence 
the  angle  is  said  to  be  measured  by  the  arc  included 
by  its  sides.    Thus,  the  angle  COA  is  measured  by 

tlie  arc  ac,  i.  e.,  it  is  the  same  part  of  a  right  angle  that  arc  ac  is  of  arc  ad. 
(See  Trigonometry,  3-10.) 


Fig.  36. 


61.  Tlieof^em, — The  relative  lengths  of  arcs  described  luith  the 
same  radius  can  he  found  in  a  manner  altogether  similar  to  that 
give?i  in  {36)  for  comparing  straight  lines. 

III. — If  I  wish  tp  compare  the  two  arcs  db  and  cd  described  with  tlie  sama 
rfidii,  I  take  the  dividers,  and  placing  the  sharp  point  on 
d  (one  end  of  the  shorter  arc),  open  them  till  the  other 
point  is  at  c.  I  then  measure  this  distance  off  on  ab  as 
many  times  as  I  can, — in  this  case  2  times,  with  a  remain- 
der/&.  This  remainder,  fb,  I  measure  off  in  the  same 
way  upon  f?c,  and  find  it  goes  once  with  a  remainder  gc. 
This  remainder,  gc,  I  apply  to  the  arc/6,  and  find  it  goes 
once  with  a  remainder  ?ib.  This  last  remainder  I  find  is 
contained  in  the  last  preceding,  gc,  2  times.  Then,  count- 
ing up  the  parts,  I  find  that  dc  is  made  up  of  5  parts  each 
equal  to  ?ib,  and  ab  of  13  such  parts.  Therefore,  nb  is  2} 
times  as  long  as  dc.     [The  angle  0  is  therefore  2f  times  the 

angle  C] 

FiQ.  37. 


ELEMENTARY  GEOMETRY. 


Ex.  1.  Draw  an  acute  angle  and  also  an  obtuse  angle,  and  then 
compare  them  as  above. 

Ex.  2.  Draw  a  small  acute  angle  and  a  large  acute  one,  and  then 
compare  them  as  above. 

Ex.  3.  Draw  a  small  acute  angle,  and  then  draw  another  angle 
3  times  as  large. 

Ex.  4.  Draw  an  acute  angle,  and  also  a  right  angle,  and  com- 
pare them  as  above. 

SuG.— Article  {39)  shows  how  to  draw  a  right  angle. 

Ex.  5.  Draw  any  angle,  and  then  draw  another  equal  to  it. 

Ex.  6.  Show  that  the  angles  or,  b,  and  c  are  respectively  ^,  |,  and 
.6  of  a  right  angle.* 


Fig.  m. 


Fig.  38. 


Ex.  7.  Show  that  angles  a  and  b.  Fig.  39,  are  respectively  1|  and 
1^  times  a  right  angle. 

Ex.  8.  Draw  a  regular  inscribed  hexagon,  as  in  Fig.  31,  and  then 
comparing  any  one  of  its  angles  with  a  right  angle,  find  that  it  is 
1^  times  a  right  angle. 

Ex.  9.  Draw  an  equilateral  triangle,  as 
in  Fig.  31,  and  find  that  any  angle  of  it 
is  I  of  a  right  angle. 

Ex.   10.    Show  that  a  right  angle  is 
^--    measured  by  ^^  of  a  circumfei-ence. 

Solution.— If  CD  is  perpendicular  to  AB, 
the  four  angles  fomied  are  equal,  and  each  is  a 
right  angle.  But,  as  all  of  them  taken  together 
are  measured  by  the  whole  circumference,  one 
of  them  is  measured  by  i  of  the  circumference. 


*  Of  course,  absolnte  accuracy  is  not  to  be  expected  in  such  eolations. 


ABOUT  ANGLES. 


27 


62,  An  Inscribed  Angle  is  an  angle  whose  vertex  is  in 
the  circumference  of  a  circle,  and  whose  sides  are  chords,  as  A, 
Fig,  41. 

63.  Theorem.. — An  inscribed  angle  is  measured  hy  one-half  the 
arc  included  letween  its  sides. 

III. — The  meaning  of  this  is  that  an  inscribed  angle  like  A,  which  includes 
any  particular  arc,  as  cd,  is  only  half  as  large  as  an  angle  would  be  at  the  centre, 
as  cOd,  whose  sides  included  the  same  arc,  cd,  or  an  equal  arc.  Thus,  in  this 
case,  drawing  the  arc  ah  from  A  as  a  centre,  with  the  same  radius,  O^Z,  as  cd  is 
drawn  with,  I  find  that  ah  which  measures  A  is  ^  of  c^  which  measures  cOd. 


Fig.  42. 

Ex.  1.  Which  of  the  angles  a,  I,  c,  d,  e  is  the  largest  ?  What  is  a 
measured  by?    What^?    Whatc?     WhattZ?    Whate?     i^/^.  42. 

Ex.  2.  Which  is  the  greatest  angle,  a,  d,  or  c,  Fig.  43  ?  By  what 
is  a  measured  ?  By  what  Z*  ?  By  what  c  ?  What  is  the  measure 
of  a  right  angle  ?     [See  Example  10  in  the  preceding  set] 


Fig.  43. 


Fig.  44. 


Ex.  3.  Suppose  I  take  a  square  card  like  CEDF,  with  a  hole  in  one 
corner  as  at  c,  and  sticking  two  pins  firmly  in  my  paper,  as  at  A  and 
B,,  place  the  corner  of  the  card  between  them,  as  in  Fig.  44,  and 
then,  keeping  the  sides  of  the  card  snug  against  the  pins,  put  a 


28  ELEMENTARY  GEOMETRY. 

pencil  through  the  hole  C  and  move  it  around  to  A  and  then  back 
to  B  ;  what  kind  of  a  line  will  the  pencil  trace  ?  Will  it  make  any 
difference  whether  c  is  a  right  angle  or  not  ?  If  any  difference, 
what? 

Ex.  4.  By  what  part  of  a  circumference  is  an  angle  of  a  regular 
inscribed  hexagon  measured  ?  See  {So),  and  Fig.  31.  How  many 
right  angles  is  the  angle  of  the  hexagon  equal  to  ?  ^Miat  is  the 
sum  of  the  six  angles  equal  to  ?  Ans.  to  last,  8  right  angles. 

Ex.  5.  Show,  from  the  way  in  which  an  equilateral  triangle  is 
constructed  in  Fig.  31,  that  one  of  its  angles  is  measured  by  J-  of  a 
circumference,  and  hence  is  f  of  a  right  angle. 


64:,  Hieorein. —  When  two  lines  intersect,  they  form  either  four 
right  angles,  or  two  equal  acute  and  two  equal  obtuse  angles. 

III. — [The  pupil  can  illustrate  this  for  himself  by  drawing  lines  and  noticing 
what  angles  are  equal.] 

Ex.  1.  Having  a  carpenters  square,  an  instrument  represented  by 
MON,  I  wish  to  test  the  angle  O  and  ascer- 
^  tain  whether  it  is,  as  it  should  be,  a  right 

angle.     I  draw  an  indefinite  right  line  AB, 
-]        and  placing  the  angle  0  at  some  point  c  on 


Q  ^      this  line  with  ON  extending  to  the  right  on 

^:|:'^  CB,  I  draw  a  line  along  OM.     Turning  the 

|i  square  over  so  that  ON  shall  lie  on  CA,  I 

I  draw  another  line  along  OM.     Three  cases 

1 may  occur. — 1st.    Suppose    the    first    line 

^  ^  ^    drawn  along  OM  is  CF,  and  the  second  CE  ; 

what  kind  of  an  angle  is  0  ?  2d.  Suppose 
the  first  line  drawn  is  CE  and  the  second  CF  ;  what  kind  of  an  angle 
is  0  ?  3d.  Suppose  the  first  and  second  lines  drawn  along  OM  coin- 
cide and  are  CD  ;  what  kind  of  an  angle  is  O  ? 

Ex.  2.  Show  that  the  sum  of  all  the  angles  formed  by  drawing 
lines  on  one  side  of  a  given  line,  and  to  the  same  point  in  the  line, 
is  two  right  angles. 


S5,  I^rob, — To  bisect  a  given  angle. 

Solution. — I  wish  to  divide  the  angle  AOB  into  two  equal  parts,  t.  e.,  to 


ABOUT  ANGLES. 


29 


[•adius,  as  0«,  I 


Fig.  46. 


bisect  it.  With  0,  the  vertex,  as  a  centre,  and  any  convenient 
strike  an  arc,  as  ba,  cutting  the  sides  of  the  angle. 
Then  from  a  and  h  as  centres,  "with  the  same  radius 
in  each  case,  I  strilie  two  arcs  intersecting  as  at  P. 
Drawing  a  line  through  P  and  0,  it  bisects  the 
angle;  i.  e.,  the  angle  POA  =  angle  BOP.  [Let 
the  pupil  try  this  by  cutting  out  the  angle  AOB, 
and  then  folding  the  paper  along  the  line  P,  or  cut- 
ting it  through  in  the  line  OP,  and  then  putting  one 
angle  on  the  other,  and  thus  see  if  they  do  not  fit.] 

Ex.  1.  Draw  an  angle  equal  to  \  of  a  right  angle. 

SuG. — First  draw  a  right  angle  and  then  bisect  it 

Ex.  2.  Draw  an  angle  equal  to  ^  of  a  right  angle. 

SuG. — Draw  a  circle.  Inscribe  an  equilateral  triangle.  [Do  it  neatly,  by 
rule,  as  in  {55).]  Then  bisect  any  angle  of  this  triangle.  This  will  be  ^  of  a 
right  angle,  since  the  whole  angle  is  |.     See  Ex.  9  {61). 

Ex.  3.  How  does  it  appear  that  the  angle  EOF,  Fig.  31,  is  J  of  a 
right  angle  ? 


GO.  Parallel  Straight  Lines  are  such  as,  lying  in  the  same 
plane,  will  not  meet  how  far  soever  they  are  produced  either  way. 

III. — The  sides  of  this  page  are  parallel  lines,  ■ 

as  are  also  the  top  and  bottom.     The  lines  in        

Fig.  47  are  parallel.  


67.  Proh.—To  drato  a  line  through  ^'^-  ^^• 

a  given  point  and  parallel  to  a  given  line. 

r 

Solution. — I  wish  to  draw  a  line  through  the  point  O  and  parallel  to  the 
line  AB.     [The  pupil  should  first  draw  some 

line,  as  AB,  and  mark  some  point,  as  O.]     Ic— i ^ D 

take  O  as  a  centre,  and  with  a  radius  *  greater  \  \ 

than  the  shortest  distance  to  AB,  as  Oa,  draw  an  ^ 
indefinite  arc  aP.  Then  with  a  as  a  centre,  and 
tlie  same  radius,  I  draw  an  arc  from  O  to  the 
line  AB  at  h.  Taking  the  distance  Oh  (the  chord)  in  the  dividers,  I  put  the  sharp 
point  on  a  and  strike  a  small  arc  intersecting  this  indefinite  arc,  as  at  P.  Fi- 
nally, drawing  a  line  through  O  and  P,  it  is  the  parallel  sought. 

*  This  means  "  put  the  sharp  point  of  the  dividers  on  O  and  open  them  till  the  distance  be- 
tween the  points  (the  radiu?)  is  more  than  the  distance  from  0  to  AB-" 


■B 


Fig.  48. 


30 


ELEMENTARY  GEOMETEY. 


68,  Theorem, — Tioo parallel  lines  are  everyiuhere  the  same  dis- 
tance apart. 

III. — Let  AB  and  CD  be  two  parallel  lines.    I  will  examine  them  at  the  two 

points  O  and  P.     To  find  how  for  apart  the 
—  B  lines  are  at  these  points  I  draw  the  perpen- 

diculars OM  and  PN.     [The  pupil  should  not 

f^    D    guess  at  these,  but  actually  draw  them  as  in- 

FiG.  49.  structed  in  (^4).]  Measuring  these,  I  find  them 

equal. 

We  can  understand  that  this  proposition  must  be  true,  since  the  lines  could 

not  approach  each  other  for  awhile  and  then  separate  more  and  more  without 

being  crooked ;  or,  if  they  kept  on  approaching  each  other,  they  would  meet 

after  awhile,  and  so  not  be  parallel. 


M 


69,  Hieorem, — Parallel  lines  mahe  no  angle  luith  each  other. 

III. — Let  AB  be  a  straight  line,  and  suppose   CD  another  sti-aight  line 

passing  through  the  point  O.  Now  let 
CD  turn  around,  first  into  the  position 
D'C,  then  into  D"C",  etc.,  all  the  time 
passing  through  O.  It  is  evident  that 
the  angle  which  this  line  makes  with 
the  line  AB  is  all  the  time  growing  less, 
i.  e.,  a'  <  a,  and  a"  <  a'.  It  is  also  evi- 
dent that  this  angle  will  become  0 
when  the  lines  become  parallel ;  for  it 

becomes  less  and  less  all  the  time,  but  is  always  somethuig  so  long  as  the  lines 

are  not  parallel. 


70,  Hieorem, — Parallel  lines  have  the  same  direction  with 
each  other. 

III.— Thus,  in  Fig.  47,  the  parallel  lines  all  extend  to  the  right  and  left,  i.  e., 
in  the  same  direction. 

Ex.  1.  How  shall  the  farmer  tell  whether  the  opposite  sides  of 
his  farm  are  parallel  ? 

Ex.  2.  If  we  wish  to  cross  over  from  one  parallel  road  to  another, 
is  it  of  any  use  to  travel  farther  in  the  hope  that  the  distance  across 
will  be  less  ? 

Ex.  3.  If  a  straight  line  intersects  two  parallel  lines,  how  many- 
angles  are  formed?  How  many  angles  of  the  same  size?  May 
they  all  be  of  the  same  size  ?  "When  ?  When  will  they  not  be  all 
of  the  same  size  ? 


ABOUT  TRIANGLES. 


31 


SECTION  IV. 

ABOUT   TRIANGLES. 

71.  A  Plane  Triangle,  or  simply  A  Triangle,  is  a  plane 
figure  bounded  by  three  straight  lines. 

7^.  With  respect  to  their  sides,  triangles  are 
distinguished  as  Scalene,  Isosceles,  and  Equilateral. 
A  scalene  triangle  has  no  two  sides  equal.  An 
isosceles  triangle  lias  two  sides  equal.  An  equi- 
lateral triangle  has  all  its  sides  equal. 

73,  With  respect  to  their  angles,  triangles  are 
distinguished  as  acitte  angled,  right  angled,  and 
obtuse  angled.  An  acute  angled  triangle  has  three 
acute  angles.  A  right  angled  triangle  has  one  right 
angle,  and  the  side  opposite  the  right  angle  is  called 
the  hypotenuse.  An  obtuse  angled  triangle  has  one 
obtuse  angle. 

Ex.  Fig.  51  affords  illustrations  of  all  the  different  kinds  of 
triangles.  Let  the  pupil  point  them  out  until  he  is  perfectly  familiar 
with  the  terms.  He  should  also  practise  drawing  the  different  kinds 
of  triangles,  for  the  purpose  of  familiarizing  the  names  applied  to 
the  different  kinds. 


Fig.  5L 


74.  Theorem, — The  sum  of  the  angles  of  a  triangle  is  tioo 
right  angles. 

<r  III. — Cut  out  any  triangle  from  a  piece  of  paper. 
Then  cut  off  two  of  the  angles,  as  1  and  2,  and  turn 
them  about  and  place  them  by  the  side  of  the  other 
angle,  as  in  the  lower  figure.  You  will  then  see  that 
the  line  OP  is  straight,  and  that  the  three  angles  of 
the  triangle  just  make  up  the  two  right  angles  OED 
and  PED. 

Ex.  1.  If  one  angle  of  a  triangle  is  a  right 
angle,  what  is  the  sum  of  the  other  two  ? 

Ex.  2.   Can  a  triangle  have  more  than  one 
right  angle  ?     If  two  of  its  angles  were  right  angles,  what  would 
the  third  angle  be  ? 


o2 


ELEMENTARY  GEOMETRY. 


Ex.  3.    Can  a  triangle  have  more  than  one  obtuse  angle  ? 

SuG. — Tiy  and  see  if  you  can  di-aw  a  triangle  -with  two  right  angles,  or  two 
obtuse  angles. 

Ex.  4.  Construct  any  triangle,  and  draw 
arcs  measuring  its  angles.  Then  draw  a  circle 
with  the  same  radius  as  the  one  used  to 
measure  the  angles,  and  lay  off  upon  the  cir- 
cumference the  arcs  measuring  the  angles. 
The  sum  of  these  arcs  will  always  make  up 
just  a  semi-circumference.  What  does  this 
show  ? 

Fig.  53.  j]x.  5.   If  two  angles  of  one  triangle  are 

equal  to  two  angles  of  another,  can  the  third  angles  be  unequal  ? 
Why? 


75,  Prob, — To  maJce  tioo  triangles  jiist  alike. 


Solution. — There  are  three  ways  of  doing  this : 

\Ht  Way. — Suppose  I  have  any  triangle,  as  ABC, 
and  want  to  make  another  just  like  it.  I  first  draw 
an  arc  measuring  any  one  of  the  angles,  as  A,  of  the 
given  triangle.  Then  I  make  an  angle  D  equal  to 
the  angle  A,  and  draw  the  sides  De  and  D/.  Now  I 
measure  DE  —  AB,  and  DF=  AC.  If  I  now  draw  EF, 
the  triangle  DEF  will  be  just  like  ABC,  so  that,  were 
I  to  cut  them  out,  I  could  apply  one  like  a  pattern  to 
the  other,  and  it  would  just  fit. 

2d  Way. — I  have  a  triangle  A,  and  wish  to  make 
another  just  like  it.  I  draw  arcs  measuring  any  hoo 
of  its  angles,  as  0  and  P.  Then,  making  a  line  MN 
equal  to  OP,  I  make  an  angle  at  M  equal  to  O,  and 
one  at  N,  on  the  same  side  of  MN,  equal  to  P. 
Now  making  these  two  sides  M6  and  Na  long  enough 
to  meet  (or,  as  we  say,  "  producing  them  till  they 
meet"),  I  have  a  second  triangle,  B,  just  like  the  first 
triangle,  A.  Were  I  to  cut  out  the  first  triangle,  it 
would  fit  on  the  second  just  like  a  pattern. 

M  Way. — I  have  a  triangle  ACB,  and  want  to  make 
another  just  like  it.  I  make  a  line  DE  equal  to  some 
side  of  the  given  triangle,  as  AB.  Then  taking  AC  as 
radius,  I  describe  an  arc  from  D  as  a  centre,  and  in 
like  manner,  with  BC  as  radius  and  E  as  a  centre. 


ABOUT  TRIANGLES. 


33 


describe  another  arc.  Through  the  intersec- 
tions of  these  arcs,  as  F,  I  draw  DF  and  EF, 
The  triangle  DEF  is  just  lil^e  ABC.  [Tiy  it  by 
drawing  as  described,  and  then  cutting  out  one 
triangle,  and  seeing  if  you  cannot  fit  it  as  a 
pattern  on  the  other.] 

Ex.  1.  In  any  triangle,  which  side  is 
opposite  the  greatest  angle  ?  Which  op- 
posite the  least  angle? 


Tig.  50. 


Ex.  2.  If  you  have  two  triangles  Avith  an  angle  in  each  equal,  hut 
the  sides  about  this  angle  longer  in  one  triangle  than  in  the  other, 
can  you  make  one  fit  on  the  other  as  a  pattern  ?  Cut  out  two  such 
triangles  and  try  it. 

Ex.  3.  Can  you  make  a  triangle  so  that  one  of  its  sides  shall  be 
as  long  as  both  the  others,  or  longer  than  both  ? 

Ex.  4.  Can  you  make  a  triangle  so  that  one  of  its  sides  shall  be 
less  than  the  difference  between  the  other  tAvo,  or  equal  to  the 
difference  ? 

Ex.  5.  If  you  have  two  triangles  with  only  one  side  ancl  one  angle 
in  the  one  equal  to  one  side  and  one  angle  in  the  other,  can  you 
apply  one  as  a  pattern  and  make  it  fit  on  the  other  ?  Cut  out  two 
such  triangles  and  try  it. 

Ex.  6.  If  you  have  two  triangles  with  only  two  sides  of  one  re- 
spectively equal  to  two  sides  of  the  other,  can  you  make  one  fit  as  a 
pattern  on  the  other  ?     Try  it. 

Ex.  7.  If  you  have  two  triangles  with  two  sides  in  one  equal  re- 
spectively to  two  sides  in  the  other,  and  the  included  angle  in  one 
greater  than  in  the  other,  how  is  it  with  the  third  sides  of  the 
triangles  ? 


76.  TJieorem, — Tlie  lines  whicJi  bisect  the  angles  of  a  triangle 
meet  loithin  the  tria^igle  at  a  common  ])oint. 

III.— Try  it,  by  drawing  a  triangle,  and  then  bisect- 
ing its  angles,  as  taught  in  (65).  You  will  need  to  do 
it  very  neatly,  or  the  lines  will  not  meet.  It  is  a 
delicate  operation.  Try  it  in  various  forms  of  triangles^ 
as  equilateral,  right  angled,  scalene,  obtuse  angled,  etc. 

3 


Etc.  57. 


84 


ELEMENTARY  GEOMETRY. 


TJieorem*- 


■Tlie  lines  drawn  from  the  vertices  of  a  triangle 
to  the  middle  of  the  opposite  sides  meet  in  a 
common  point  within  the  triangle, 

III. — Draw  a  triangle.  Bisect  each  of  the  sides 
as  taught  in  {39).  Tlien  join  eacli  angle  and  the 
middle  of  its  opposite  side  with  a  straight  line.  If 
you  do  the  work  well,  the  three  lines  will  cross 
each  other  at  a  common  point  within  the  tiiangle. 


Fig.  58. 


78.  Theorem* — The  perpendiculars  which  bisect  the  sides  of  a 
triangle  meet  at  a  commoyi  point,  which  may  le 
luithin  or  icithout  the  triangle,  or  m  one  of  its 
sides,  according  to  the  form  of  the  triangle. 


III. — Draw  an  acute  angled  triangle^  and  bisect  its 
sides  by  perpendiculars.  If  you  do  it  with  accuracy, 
they  will  meet  at  a  common  point  within  the  triangle. 

Draw  an  obtuse  angled  triangle,  bisect  its  sides  with 
perpendiculars,  and  they  will  meet  at  a  common 
point  without  the  triangle. 

Draw  a  right  angled  triangle,  and  the  pei*pendiculars 
will  meet  in  the  side  opposite  the  right  angle  (the 
hypotenuse). 


Ex.  1.  Draw  an  equilateral  triangle,  and  find 

the  three  points  characterized  in  the  last  three 

articles.    Are  they  all  in  one  place,  or  are  they 

Fig.  59.  in  different  places  ? 

Ex.  2.  Draw  a  scalene  triangle,  and  find  the  three  points  as  above. 

Are  they  ^11  in  the  same  place,  or  are  they  in  different  places  ? 


^^, 


79.  JProb. — To  inscribe  a  circle  in  a  given  triangle, 
C 


Fig.  60. 


Solution. — I  wish  to  inscribe  a  circle  in 
the  triangle  ABC  ;  that  is,  a  circle  to  which 
the  sides  of  the  triangle  shall  be  tangents. 
[First  draw  the  triangle.]  I  bisect  the  angles 
as  taught  in  (6*5) ;  and  then  from  the  point 
O,  where  these  intersect,  I  let  fall  perpen- 
diculars upon  the  sides,  as  taught  in  (45). 
Then  from  0  as  a  centre,  with  a  radius  equal 


ABOUT  EQUAL  FIGURES. 


35 


to  one  of  these  perpendiculars  (they  are  all  equal),  I  draw  a  circle,  and  it  is  the 
circle  required. 


SO,  Prob, — To  circumscribe  a  circle  about  a  given  triangle. 

Solution. — I  wish  to  circumscribe  a  circle 
about  the  triangle  ABC  which  I  have  drawn.  To 
do  this,  I  bisect  the  sides  with  perpendiculars,  and 
find  their  common  intersection  0,  as  taught  in 
{78).  With  0  as  a  centre  and  a  radius  equal  to 
OB,  the  distance  from  0  to  the  vertex  of  any  one 
of  the  angles,  as  these  distances  are  all  equal,  I 
draw  a  circle.  This  is  the  circumscribed  circle, 
that  is,  the  circle  in  whose  circumference  the  ver- 
tices of  the  triangle  lie.  [This  is  really  the  same 
as  Prob.  {58).]  Fig.  «i. 


SECTION   V. 

ABOUT  EQUAL  FIGURES. 

81*  Equal,  in  geometry,  signifies  alike  in  all  respects,  i.e.,  of  the 
same  shape  and  the  same  size. 

82*  Equivalent  figures  are  such  as  have  the  same  area,  i.  g.,are 

of  the  same  size,  irrespective  of  their  form. 

Ex.  1.  Can  a  triangle  be  equal  to  a  circle  ?  Can  it  be  equivalent  ? 
Can  a  circle  be  equivalent  to  a  square  ?    Can  it  be  equal  to  a  square  ? 

Ex.  2.  Can  a  right  angled  triangle  be  equal  to  an  equilateral  tri- 
angle ?  Can  a  right  angled  triangle  be  equal  to  an  isosceles  triangle  ? 
If  either  is  possible,  construct  figures  illustrating  it. 


83.  I*rob*—To  apply  one  straight  line  to  another, 

^oi^VTio^.— [Applying  ^gwre?,  to  each  other  is  a  very  important  thing  in 
geometry,  and  may  seem  a  little  curious  at  first ; 
but  it  is,  in  reality,  very  simple.  The  pupil  must 
become  perfectly  familiar  with  it.]  We  will  first 
apply  the  line  AB  to  the  equal  line  CD.  Take 
the  line  AB,*  and  placing  the  end  A  upon  the 
end  C  of  the  line  CD,  make  the  line  AB  take  the 
same  direction  as  CD,  and  put  the  former  upon 
the  latter.    Now,  since  the  lines  are  equal,  the  ^^^-  62. 

*  That  ie,  think  about  it  just  fis  if  it  were  a  little  rod  which  you  could  pick  np  and  handle. 


H 


56 


ELEMENTARY  GEOMETRY. 


extremity  (or  the  point)  B  -will  fall  upon  D,  and  the  two  lines  "will  coincide 
throughout  their  whole  extent. 

Again,  we  will  apply  the  line  EF  to  the  line  CH.  Taking  the  line  EF  {think 
of  it  as  a  little  rod  which  you  can  pick  up  and  handle),  put  the  point  E  upon 
C,  and  making  the  line  EF  take  the  same  direction  as  CH,  put  the  former  upon 
the  latter.  Now,  since  EF  is  shorter  than  CH,  the  point  (extremity)  F  will  fall 
somewhere  on  the  line  CH,  as  at  I.  Therefore  the  lines  do  not  coincide 
throughout  their  whole  extent,  and  are  not  equal. 


84.  Proh, — To  apply  one  plane  angle  fo  another. 

Solution. — First  we  will  apply  one  angle  to  another  equal  angle.  Thus,  to 
apply  BAC  to  the  equal  angle  EDF.  Take  the 
angle  BAC  {think  of  it  as  if  it  were  two  little  rods 
put  firmly  together  at  this  angle,  and  so  that  you 
could  pick  tliem  up  and  handle  them),  and  placing 
the  vertex  (point)  A  upon  the  vertex  (j^oint)  D^ 
make  the  side  AC  take  the  direction  DF.  As 
AC  happens  to  be  longer  than  DF,  the  extremit}- 
C  will  fall  beyond  F,  at  some  point,  as  0.  But  we 
do  not  care  for  this,  as  the  size  of  an  angle  does 
not  depend  upon  the  length  of  the  sides.  Now, 
while  A  lies  on  D,  and  the  line  AC  on  DF,  let  the 
line  AB  be  conceived  as  lying  in  the  plane  of  the 
paper  also  {i.  e.,  on  it).  Since  the  angle  BAC  is 
equal  to  EDF,  the  line  AB  will  take  the  direction 
DE,  and  will  fall  on  it,  though  the  point  B  will 
fall  somewhere  beyond  E,  as  at  N,  as  AB 
chances  to  be  longer  than  DE.  The  two  angles 
therefore  comcide,  and  are  equal.  [Notice  care- 
fully just  what  is  meant  by  saying  tliat  the  angles 
are  equal.  We  do  not  mean  that  the  sides  are 
of  the  same  length,  but  that  the  opening  between 
them  is  the  same,  i.  e.,  that  one  is  just  as  sharp 
a  comer  as  the  other.] 

Queiies. — If  BAC  were  greater  than  EDF,  and 
we  should  begin  by  putting  A  upon  D,  and  make  AC  fiill  upon  DF,  where  would 
AB  fall,  without  the  angle  EDF  or  within  it?  If  BAC  were  less  than  EDF,  and 
we  proceed  as  before,  placing  the  vertex  A  on  D,  and  AC  on  DF,  would  AB  fall 
without  EDF  or  within  it  ? 

Again,  let  us  attempt  to  apply  the  angle  HGI  to  LKM.  Placing  the  vertex  C 
on  the  vertex  K,  making  the  side  CI  take  the  direction  KM,  and  then  bringing 
CH  into  the  plane  of  the  paper,  the  side  CH  will  f^ill  within  the  angle  LKM  (aa 
in  the  line  KR),  since  the  angle  HCI  is  less  than  LKM.  The  angles,  tlierefore, 
do  not  coincide. 


ABOUT  EQUAL  FIGURES.  37 

8S,  JProb* —  When  two  triangles  have  two  sides  and  the  included 
angle  of  one  equal  to  iiuo  sides  and  the  included  angle  of  the  other,  to 
apply  one  triangle  to  the  other. 

Solution. — In  the  two  triangles  ABC  and  DEF,  lef  the  angle  A  be  equal  to 
the  angle  D,  the  side  AB  =  the  side  DE,  and  AC  =  DF.  We 
will  apply  tlie  triangle  ABC*  to  DEF.  Take  the  triangle 
ABC  and  place  the  vertex  A  upon  the  vertex  D,  making  the 
side  AC  take  the  direction  DF.  Since  AC  =  DF,  the  ex- 
tremity C  will  then  fall  on  F.f  Now  bring  the  triangle  ABC 
into  the  plane  of  DEF,  keeping  AC  in  DF,  and  the  line  AB 
will  take  the  direction  DE,  since  the  angle  A  =  the  angle  D. 
Again,  as  AB  =  DE,  the  extremity  B  will  fall  upon  E.  Thus 
we  have  placed  ABC  upon  DEF,  so  that  A  falls  upon  D, 
C  upon  F,  and  B  upon  E,  and  find  that  they  exactly 
coincide. 

Fig.  64. 

Ex.  1.  Suppose  you  attempt  to  apply  ABC  in  the  last  figure  to 
DEF  by  placing  B  on  D,  and  letting  BC  fall  upon  DF.  Where  will  C 
fall  ?  Measure  it  and  find  out.  Which  side  will  then  fall  nearly  or 
quite  on  DE  ?  Will  it  fall  exactly  on  it  ?  On  which  side  will  it  fall  ? 
Can  you  make  the  triangles  coincide  (fit)  in  this  way  ? 

Ex.  2.  Can  you  make  the  triangles  in  the  last  figure  coincide  by 
placing  C  upon  D,  and  letting  CA  fall  upon  DF  ?  Where  will  a  fall  ? 
What  line  will  fall  on  or  near  DE  ?  Will  it  fall  without  DE,  or 
within  ? 

Ex.  3.  Construct  two  isosceles  triangles,];  as  ACB 
which  AC  =  CB  =  DE  =  EF.  Can  you  ap- 
ply DEF  to  ABC  by  putting  D  upon  A  ? 
Describe  the  process.  Can  you  put  D 
upon  A  and  DE  upon  AB,  and  make  the 
triangles  coincide?  Can  you  make  the 
triangles  coincide  by  putting  F  upon  A  ? 
If  so,  describe  the  process.  Can  you  make  them  coincide  by  putting 
E  upon  A  ?     If  not,  point  out  the  difficulties. 


*  Think  of  ABC  as  made  of  little  rods,  eo  that  you  can  pick  it  up  and  place  it  upon  DEF- 
in  the  manner  described. 

t  It  will  make  it  clearer  if  the  pupil  think?  of  ABC.  at  this  stage  of  the  operation,  as 
having  the  side  AC  on  DF,  but  the  angle  B  not  down  on  the  paper ;  just  as  if  he  were  to  cut 
out  ABCi  and  set  the  edge  AC  on  the  line  DF,  and  afterward  bring  the  triangle  ABC  down 
on  to  DEF,  keeping  the  edge  AC  on  the  line  DF- 

X  The  teacher  must  insist  upon  the  figures  being  drawn,  and  that  accurately,  according  to 
rule. 


38 


ELEMENTABY  GEOMETRY. 


Ex.  4.  Construct  two  equal  trapeziums,* 
as  ABCD  and  EFCH,  and  describe  the  process 
of  applying  one  to  the  other. 

Solution.— I  will  apply  EFCH  to  ABCD.  As 
the  angle  E  is  equal  to  the  angle  B,  I  will  begin  by 
putting  the  vertex  E  on  B,  and  making  EH  fall  upon 
BC.  Since  EH  =  BC,  H  will  fall  on  C.  Now,  as 
angle  H  =  angle  C,  HC  Tvill  take  the  direction  (fall 
on)  CD;  and  since  HC  =  CD,  C  will  fall  on  D. 
Again,  as  C  =  D,  CF  will  take  the  direction  DA; 
and  since  CF  =  DA,  F  will  fall  on  A.  Finally,  as 
F  =  A,  FE  will  take  the  direction  AB;   and  since 

FE  =  AB,  E  will  fall  on  B,  as  it  ought,  since  I  started  by  conceiving  E  as 

placed  on  B. 

Ex.  5.  Describe  the  application  of  ABCD  in  the  last  figure  to  EFCH, 
by  beginning  with  C  upon  H. 

Ex.  6.  Having  two  equal  equilateral  triangles,  can  you  apply  one 
to  the  other  by  beginning  indifferently  with  any  one  angle  of  one 
upon  any  one  angle  of  the  other  ?  Draw  two  such  triangles,  and  go 
through  with  the  details  of  the  application. 


86 •  JProb, — Given  huo  triangles  with  two  angles  and  the  included 
side  of  the  one  respectively  equal  to  two  angles  and  the  included  side 
of  the  other,  to  apply  one  triangle  to  the  other. 

Solution. — [The  pupil  should  first  draw  any  triangle,  as 
ABC.  Then  make  a  line  DF  equal  to  AB,  and  at  the  ex- 
tremities D  and  F  make  angles,  as  D  and  F,  respectively 
equal  to  A  and  B.  This  is  preliminary.]  Having  the  two 
triangles  ABC  and  DEF,  in  which  A  =  D,  B  =  F,  and  AB  =  DF, 
I  propose  to  apply  one  to  the  other.  I  will  apply  ABC  to 
DEF.  Taking  ABC,  I  place  A  upon  D,  and  make  AB  take 
the  direction  and  fall  upon  DF.  Since  AB  =  DF,  B  will  fall 
upon  F.  Now  keeping  the  line  AB  in  DF,  I  conceive  the 
triangle  ABC  to  come  into  the  plane  of  DEF.  Since  A  =  D, 
tlie  side  AC  will  take  the  direction  DE,  and  the  extremity  C 
of  AC  will  fall  somewhere  in  the  line  DE,or  in  DE  produced, 
the  line  BC  will  take  the  direction  FE,  and  the  extremity  C 


Fig.  67. 
Also,  since  B  =  F. 


of  BC  will  fall  somewhere  in  FE  or  FE  produced.   Finally,  as  C  falls  in  DE  and 


*  Thp  teacher  must  insist  upon  the  figures  being  drawn,  and  that  accurately,  according  to 
»ule. 


ABOUT  SIMILAR   FIGURES,  ESPECIALLY  TRIANGLES. 


39 


FE  both,  it  must  be  at  E,  their  intersection.    Thus  I  find  that  the  triangle  ABC, 
when  applied  to  DEF,  coincides  with  it  throughout.. 

Ex.  1.  Given  the  two  triangles  DEF  and  ABC,  in  Avhich  DE=:AB, 
D  =  A,  but  E  >  B  ;  show  how  an  attempt  ^/f 

to  apply  one  to  the  other  fails. 

Solution. — Since  angle  D  =  angle  A  *  I  ap- 
ply the  vertex  D  to  the  vertex  A,  and  make  DE 
take  the  direction  AB.  AsDE^rAB,  E  will  fall 
on  B,  and  the  sides  DE  and  AB  will  coincide. 
Again,  since  D  =  A,  the  side  DF  will  take  the 
direction  AC  when  the  planes  of  the  triangles 
coincide ;  and  the  extremity  F  will  fall  in  AC, 
or  in  AC  produced  (really  in  AC  produced,  in 
this  case).  Finally,  since  E  >  B,  EF  will  fall  to 
the  right  of  BC,  and  the  application  fails.  yig.  G8. 

Ex.  2.  Construct  two  trapeziums  with  their  respective  sides  equal, 
as  AC  =  HE,  AB  =  HG,  BD  =:  OF,  and  CD  =  EF, 
but  with  their  angles  unequal ;  and  show  how 
an  attempt  to  apply  one  to  the  other  fails. 

Ex.  3.  If  the  sides  of  two  trapeziums,  as  in 
the  last  figure,  are  equal,  and  two  of  the 
angles  including  a  side  in  one  are  respectively 
equal  to  the  corresponding  angles  in  the  other, 
as  A  =  H,  and  B  =  C,  can  one  be  applied  to  the 
other  ?    If  so,  give  the  details  of  the  process.  fig.  c9. 


SECTION  VL 

ABOUT  SIMILAR  FIGURES,  ESPECIALLY  TRIANGLES. 

87,  Similar  Figures  are  such  as  are  shaped  alike — i.  e.,  have 
the  same  form. 

A  more  scientific  definition  is.  Similar  Figures  are  such  as  have 
their  angles  respectively  equal,  and  their  homologous  (correspond- 
ing) sides  proportional. 


*  Be  careful  to  distinguish  betweea  the  vertex,  which  is  a  point,  and  the  angle,  which  is  the 
opening  between  the  lines. 


^ 


ELEMENTARY  GEOMETRY. 


Fig.  70. 


88.  Honioloffous^  or  Corresponding  Sides  of  similar  figures, 
are  those  which  are  included  between  equal  angles  in  the  respective 
figures. 

In  Similar  Triaxgles,  the  Homologous  Sides  are  those 

OPPOSITE  THE   equal  AJh^GLES. 

III. — The  triangles  ABC  and  DEF  are  similar,  for  they  are  of  the  same 

shape.  But  it  is  easy  to  see  that 
ABC  is  not  similar  to  IHK  or 
MON.  The  pupil  should  notice 
that  A  =  D,  C  r=  F,  and  B  =  E. 
Also,  side  e  is  IJ  times  5,  side  /is 
1^  times  c,  and  side  rf  is  1^  times 
a;  so  that  f  :  c  :  :  e  :b,  and 
/  :  c  : :  d  :  a,  and  d  \a  :  :  e  :  b. 
Now  there  are  no  -such  relations 
existing  between  the  parts  of 
ABC  and  IHK.  The  angles  B 
and  K  are  nearly  equal,  but  A  is 
much  larger  than  H,  and  C  is 
smaller  than  I.  So  these  triangles  are  not  mutually  equiangular,  i.  e.,  each  angle 
in  one  has  not  an  equal  angle  in  the  other.  Again,  as  to  their  sides,  IH  is  a 
little  less  than  AC,  but  H  K  is  greater  than  AB.  These  two  triangles  are,  there- 
fore, not  similar. 

In  the  similar  triangles  ABC  and  DEF,  6  is  homologous  with  e,  since  they  are 
opposite  the  equal  angles  B  and  E.  For  a  like  reason  a  is  homologous  with  dy 
and  c  with  f.  It  may  also  be  observed,  that  the  shortest  sides  in  two  similar 
triangles  are  homologous  with  each  other ;  the  longest  sides  are  also  homolo- 
gous with  each  other,  and  the  sides  intermediate  in  length  are  homologous 
with  each  other. 

Ex.  1.  Can  a  scalene  triangle  be  similar  to  an  isosceles  triangle  ? 
Can  an  obtuse  angled  triangle  be  similar  to  a  right  angled  triangle  ? 

Ex.  ^.  Are  all  squares  similar  figures  ? 

SuG. — First,  are  the  angles  equal?  Second,  is  any  one  side  of  one  square  to 
some  side  of  another  square  as  a  second  side  of  the  fii*st  is  to  a  second  side  of  the 
second,  etc.  ? 

Ex.  3.  A  farmer  has  two  fields,  each  of  which  has  4  sides  and  4 
right  angles.  The  first  field  is  20  rods  by  50,  and  the  other  40  by 
80.     Are  they  similar  ? 

SuG.— Are  they  mutually  equiangular?  Then  are  the  lengths  in  the  same 
ratio  as  the  widths?  If  they  are  not  similar,  how  long  would  the  second  have 
to  be  in  order  to  make  them  similar?  Draw  two  such  ligures,  and  see  if  they 
look  alike  in  shape- 


ABOUT  SIMILAR  FIGURES,   ESPECIALLY  TRIANGLES.  41 

89.  I*roh, — To  find  a  fourth  proportional  to  three  given  lines. 


Solution. — I  have  the  three  given  lines 
A,  B,  and  C,  and  wish  to  find  a  fourth  line      ^ 
such  that  Q 

A  shall  be  to  B  as  C  is  to  X]ie  fourth  Um,  i.  d., 
A  :  B  :  :  C  :  fourth  line. 


To  do  this,  I  draw  two  indefinite  lines  OX 
and  OY,  from  a  common  point  0.  On  one 
of  these,  as  OX,  I  lay  off"  Oa  =  A,  and  Oc  = 
B.  Then  on  the  other  1  make  Ob  =  0,  and 
draw  ab.  Finally,  drawing  a  parallel  to  ab 
through  the  point  c  (67),  I  have  Od  as  the  line 
sought.    Thus,  calling  OfZ,  D,  the  proportion  is 


Fig.  71. 


Oa     :     Oc 
A     :     B 


Ob 
0 


Od.OT 
D. 


N.B. — The  order  in  which  the  lines  are  taken,  and  the  loay  of  drawing  the  lines 
ab  and  cd,  are  essential.  The  following  directions  will  insure  correctness  :  Lay 
off  the  FIRST  a?id  second  on  the  same  line,  as  on  OX ;  and  the  third  on  the 
OTHER  LINE,  as  on  OY.  Then  join  the  extremities  of  the  first  and  third,  and 
draw  tlie  parallel  through  the  extremity  of  the  second. 


Fig. 


Ex.  1.  Show  that  if  the  order  of  the  proportionals  in  Fig.  71  is 
B  :  A  : :  C  :  fourth  line,   the   fourth 
proportional  is  e,  Fig,  71. 

Ex.  2.  Show  that  a  fourth  propor- 
tional to  A,  B,  and  C  is  D.  Also,  that 
a  fourth  proportional  to  C,  A,  and  B 
is.E.  Show  that,  if  the  order  he 
A  :  C  : :  B  :  fourth  line,  D  is  still  the 
fourth  proportional.  Show  that 
B  :  C  : :  A  :  2C,  nearly. 

Ex.  3.  Solve  the  proportion  3  :  8  : :  5  :  a;,  and  find  x  geometrically. 


SuG.— Using  the  scale  of  lOOths  of  a 
foot,  the  figure  is  that  in  the  margin.  OD 
is  the  foiu'th  proportional,  or  x  =  OD, 
which  is  found  bi/  measurement  to  be  10  j, 
as  it  is  by  arithmetic. 


Fio.  73. 


42 


ELEMENTAEY  GEOMETRY. 


00,  I^rob, — To  draio  a  triangle  similar  to  a  given  tria?igle,  and 
having  a  given  side. 

Solution. — 1st  Method. — I  have  a  triangle  ACB,  and  want  to  make  another 
similar  to  it,  but  having  the  side  homologous  to  BC  equal  to  a.  I  draw  an 
indefinite  line,  and  on  it  take  EF,  equal  to  a.  Then  at  F  I  make  an  angle  equal 
to  C,  and  make  the  side  indefinite.  Now  I  find 
a  fourth  proportional  to  BC,  EF,  and  AC.  Having 
found  this,  as  in  the  last  article,  I  lay  it  ofi"  from 
F,  as  FD.  Drawing  DE,  I  have  DEF,  the  triangle 
required. 

I  can  readily  satisfy  myself  that  DEF  is  simi- 
lar to  ABC,  for  besides  the  foct  that  it  looks  as  if 
it  were  of  the  same  shape,  by  measuring  the  other 
two  angles,  I  find  that  E  :=  B,  and  D  =  A.  More- 
over, I  know  that  BC,  EF,  AC,  and  DF  are  pro- 
portional, because  I  made  them  so.  And,  by 
finding  a  fourth  proportional  to  BC,  EF,  and  AB, 
I  find  it  exactly  equal  to  DE.  In  like  manner 
constructing  a  fourth  proportional  to  AC,  DF,  and 
A B,  I  find  it  to  be  DE.  So  that  the  two  triangles 
are  mutually  equiangular,  and  have  their  homolo- 
gous sides  (those  opposite  the  equal  angles)  pro- 
portional.   Hence,  the  triangles  are  similar. 

2d  Method. — But  an  easier  way  to  construct  DEF  is  to  make  the  angle  F  = 
C  as  before,  and  then  make  E  =  B,  and  produce  the  sides  till  they  meet  in  D. 
The  triangles  will  then  be  similar,  and  the  proportionality  of  the  sides  can  be 
tested. 

Ex.  1.  Given  a  triangle  whose  sides  are  7,  11,  and  15,  to  construct 
a  similar  triangle  having  the  side  corresponding  to  the  one  which  is 
11  in  the  given  triangle,  8. 

Ex.  2.  Construct  two  triangles  with  equal  angles,  and  then  com- 
pare the  sides,  and  see  whether 
you  can  make  two  triangles  whose 
angles  shall  be  respectively  equal, 
and  their  sides  not  be  propor- 
tional. 


SuG. — Having  the  triangle  ABC,  make 
DEF  equiangular  with  it,  and  then 
compare  the  homologous  sides.  In  the 
figure  D  is  made  equal  to  C,  and  F  to 
A;  whence  E  =  B.  DE  and  BC  are 
homologous  sides,  because  opposite  the 
equal  angles  F  and  A.     DF  is  homolo- 


FiG.  75. 


ABOUT  SIMILAR  FIGURES,   ESPEOTATJiY  TRIANGLES. 


43 


gons  with  AC,  because  it  is  opposite  angle  E,  which  equals  B.    For  a  simi- 
lar reason.  EF  is  homologous  with  AB.    Now,  taking  two  sides  of  ABC,  as  BC 
and  AB,  and  a  side  of  DEF  homologous  with  one  of  them,  as  DE,  and  finding  a 
fourth  proportional  Oc,  it  will  be  found  exactly  equal  to  EF;  so  that 
BC  :   DE  :  :   AB  :  EF  (=  Oc). 

Ex.  3.  Make  two  triangles,  two  of  whose  angles  shall  be,  one  f 
and  the  other  J  of  a  right  angle ;  but  make  the  side  included  between 
these  angles  twice  as  great  in  the  second  triangle  as  in  the  first. 
What  will  be  the  ratio  of  the  side  opposite  the  angle  f  in  the  first 
triangle  to  the  homologous  side  in  the  second  ?  What  the  relation 
of  the  sides  opposite  the  angles  ^  ? 

Ex.  4.  If  you  make  one  triangle  whose  sides  are  5,  8,  and  3 ;  and 
a  second  wiiose  sides  are  15,  24,  and  9,  will  they  be  mutually  equi- 
angular? Which  angles  are  the  equal  ones  ?  W^hich  are  the  homol- 
ogous sides? 

Ex.  5.  There  are  three  pairs  of  similar 
triangles  in  Fig.  76.  Can  you  point  them 
out?  Also  point  out  their  homologous 
parts.  Are  all  the  triangles  which  you 
can  make  out  from  the  figure  similar  to 
each  other  ? 


I'iG.  76. 


Fig.  77. 


Ex.  6.  Wishing  to  know  the  height  EC  of  a  house,  I  set  up  a 
stake  DB  5  feet  long;  and  putting  my 
eye  close  to  the  ground,  I  moved  back 
from  the  stake  to  A,  so  that  the  top  of 
the  stake  and  the  top  of  the  house  were 
just  in  range  (in  a  line).  Then  by  meas- 
uring I  found  AB  =  10  feet,  and  AC  =  80 
feet.    What  was  the  height  of  the  house  ? 

Ex.  7.  If  you  take  three  sticks  of  different  lengths  and  put  them 
together  by  joining  their  ends  two  and  two,  so  as  to  represent  a 
triangle ;  can  you,  by  putting  together  the  same  sticks  in  a  different 
order,  make  a  triangle  of  different  form  from  the  first  ?  Will  the 
angles  opposite  the  same  sticks  always  be  the  same  ? 

Ex.  8.  If  you  take  more  than  three  sticks  (say  4),  and  make  of 
them  the  boundary  of  a  figure,  by  putting  their  ends  togetlier  two 
and  two,  can  you  put  them  together  so  as  to  make  another  figure  of 
different  form  ?     Can  you  make  figures  having  different  angles  ? 

Ex.  9.  If  you   take  three  sticks,   A  3  inches  long,  B  5  inches, 


44 


ELEMENTARY  GEOMETRY. 


and  c  6  inches ;  and  also  three  other  sticks,  D  0  inches  long,  E  15 
inches,  and  F  18  inches  ;*  can  you  place  them  together  so  as  to  make 
dissimilar  triangles  ?  Will  the  corresponding  angles  of  the  two  tri- 
angles be  equal  however  you  may  arrange  the  sticks  ?  If  the  sides 
of  two  triangles  are  proportional,  will  their  angles  be  equal  and  the 
triangles  similar? 

Ex.  10.  If  you  take  four  sticks,  A  3  inches  long,  B  5  inches,  C  6 
inches,  and  K  4  inches ;  and  also  four  other  sticks,  D  9  inches  long, 
E  15  inches,  F  18  inches,  and  L  1^  inches  ;*  can  you  place  them  to- 
gether so  as  to  make  four-sided  figures  which  shall  be  dissimilar 
{i.  e.,  not  of  the  same  shape)  ?  Will  the  corresponding  angles  of  the 
two  figures  be  necessarily  equal?  If  the  sides  of  a  four-sided  figure 
are  proportional,  does  it  follow  that  the  corresponding  angles  are 
equal,  and  the  figures  similar  ? 

Ex.  11.  Why  do  the  braces  in  the  frame 
of  a  building  stiffen  it?  Is  a  four-sided 
figure  stiff?  i.  e.,  are  its  angles  incapable 
of  change  while  its  sides  remain  of  the 
same  length  ?  Can  the  angles  of  a  triangle 
be  changed  while  the  sides  remain  un- 
changed ? 


Fig.  78. 


SECTION  VIL 

ABOUT  AREAS. 

91.  A  Quadrilateral  is  a  plane  surface  inclosed  by  four 
right  lines. 

92.  There  are  three  Classes  of  quadrilaterals,  viz.,  Trapeziums, 
Trapezoids,  and  Parallelograms. 

93.  A  Trapezium  is  a  quadrilateral  which  has  no  two  of  its 
sides  parallel  to  each  other. 

94.  A  Trapezoid  is  a  quadrilateral  which  has  but  two  of  its 
sides  parallel  to  each  other. 


*  Notice  that  the  sides  are  proportional  i.  e.,  in  the  same  ratio  taken  two  and  two. 


ABOUT  AREAS. 


45 


95o  A  Parallelogram  is  a  quadrilateral  which  has  its  oppo- 
site sides  parallel. 

^6.  A  Rectangle  is  a  parallelogram  whose  angles  are  right 
angles. 

97*  A  Square  is  an  equilateral  rectangle.* 

98,  A  Rhombus  is  a  parallelogram  whose  angles  are  not  right 
angles,  and  ail  of  whose  sides  are  equal. 

99.  A  Rhomboid  is  a  parallelogram  whose  angles  are  not 
right  angles,  and  two  of  whose  sides  are  greater  than  the  other  two. 


III. — The  figures  in  the 
margin  are  all  quadrilat- 
erals. A  is  a  trapezium. 
(Why?)  B  is  a  trapezoid. 
(Why?)  C,  D,  E,  and  F  are  / 
parallelograms.  (Why  ?) 
D  and  E  are  rectangles, 
although  D  is  the  form 
usually  referred  to  by  the 
teiTa  rectangle.  So  C  is 
the  form  usually  referred 
to  when  a  parallelogram  is 
spoken  of,  without  saying 
what  kind  of  a  parallel- 
ogram. C  is  also  a  rhom- 
boid. (Why?)  E  is  a  square. 
(Why?)  F  is  a  rhombus. 
(Why  ?)  This  page  is  a 
rectangle;  so  also  are  the 
common  panes  of  glass. 


Fig.  79. 


100.  A  Diagonal  is  a  line  joining  two  angles  of  a  figure,  not 
adjacent. 

III. — In  common  language,  a  diagonal  is  a  line  running  "from  comer  to 
corner." 

Ex.  1.  To  construct  a  square, having  given  a  side;  or,  in  other 
words,  to  construct  a  square  on  a  given  line. 


*The  pupil  Phould  be  able  to  give  this  and  all  similar  definitions  at  length.   Thus,  A  Square 
is  a  surface  inclosed  by  four  equal  right  lines  making  right  angles  with  each  other. 


46 


ELEICENTARY  GEOMETRY. 


Y 
N — 


1st  Method.— Let  A  be  the  given  side.  Draw 
the  indefinite  line  OX,  and  lay  off  OM  =  A.  At 
M  erect  a  perpendicular  MY,  as  taught  in  (44). 
On  this  take  M  N  =  A.  From  N  and  0  as  centimes, 
with  a  radius  equal  to  A,  describe  arcs  iutei-sect- 
ing,  as  at  P.    Draw  NP  and  PO. 

2d  Metliod. — Let  Q  be  the  given  side.  Con- 
struct equal  angles  at  the  extremities  of  Q,  and 
produce  tlie  sides  till  they  meet,  and  one  of 
them  till  it  will  meet  another  side  of  the  square 
proposed.  With  S  as  a  centre,  and  ST  or  SR  as 
radius,  describe  a  semicircle.  Draw  RV,  and  it 
forms  a  right  angle  at  R.  The  construction  can 
now  be  finished  as  before. 

Ex.  2.  Construct  a  rhombus  whose  side 
is  2  inches,  and  one  of  whose  acute  angles 
is  I  of  a  right  angle. 

^^^-  ^^-  Ex.  3.  Construct  a  rectangle  whose  ad- 

jacent sides  are  3  and  5.* 

Ex.  4.  Construct  a  rhomboid  whose  adjacent  sides  are  3  and  7, 
and  their  included  angle  J  a  right  angle. 

Ex.  5.  How  many  diagonals  has  a  triangle?  How  many  has  a 
quadrilateral  ?  How  many  has  a  figure  with  five  sides  (a  pentagon)  ? 
Of  six?     Of  eight? 


101,  TJie  Area  of  a  surface  is  the  number  of  timos  it  contains 
some  other  surface  taken  as  a  unit  of  measure  ;  or  it  is  the  ratio  of 
one  surface  to  another  assumed  as  a  standard  of  measure. 

102*  The  Uiiit  of  Area  usually  assumed  is  a  square,  a  side  of 
which  is  some  linear  unit:  thus,  a  square  inch,  a  square  foot,  a 
square  yard,  a  square  mile,  etc.  By  these  terms  is  meant  a  square  1 
inch  on  a  side,  one  foot  on  a  side,  one  yard  on  a  side,  etc. 

The  acre  is  an  exception  to  the  general  rule  of  assuming  the 
square  on  some  linear  unit  as  the  unit  of  area,  there  being  no  linear 
unit  in  use  whose  length  is  the  side  of  a  square  acre. 

III. — The  area  of  a  board  is  the  number  of  squares  1  foot  on  a  side  which  it 
would  take  to  cover  it.  The  area  of  a  floor  may  be  spoken  of  in  square  yards, 
and  is  the  same  as  the  number  of  square  yards  of  carpeting  it  would  take  to 
cover  it. 


*  TaVe  any  convenient  unit,  ap  \  inch,  1  inch. 


ABOUT   AREAS. 


47 


103,  The  Altitude  of  a  parallelogram  is  the  distance  between 
its  opposite  sides ;  of  a  trapezoid,  it  is  the  distance  between  its  parallel 
sides;  of  a  triangle,  it  is  the  distance  from  any  vertex  to  the  side 
opposite  or  to  that  side  produced. 

±04t,  TJie  Bases  of  a  parallelogram  or  of  a  trapezoid  are  the 
sides  between  which  the  altitude  is  conceived  as  taken ;  of  a  triangle, 
it  is  the  side  to  which  the  altitude  is  perpendicular. 


III.— The  dotted  hnes  in  B,  C,  D,  and  F,  Fig.  79,  represent 
When  the  altitude  is  the  distance  between  two  parallels,  the  figure 
bases.  The  altitude  of  a  parallelogram  may 
be  reckoned  between  either  pair  of  parallel 
sides,  but  it  is  most  common  to  conceive  it  as 
the  distance  between  the  two  longer  sides. 
The  altitude  of  a  rectangle  is  the  same  as 
either  side  to  which  it  is  parallel.  A  triangle 
may  have  three  altitudes,  and  any  side  of  a 
triangle  may  be  conceived  as  its  base.  In 
Fig.  81,  AB  is  conceived  as  the  base  in  each  case,  and  CD  the  altitude. 


altitudes, 
has  two 


Ex.  What  side  of  a  triangle  must  you  conceive  as  the  base,  in 
order  that  the  altitude  shall  fall  upon  it,  and  not  upon  its  pro- 
longation ? 
a  case  ? 


From  what  angle  will  the  altitude  be  reckoned  in  such 


105.  TJieorem, — 77ie  area  of  a  rectangle  is  the  product  of  its 
two  adjacent  sides;  or,  luhat  is  the  same  thing,  the  product  of  its 
altiticde  and  base. 


III. — Let  ABCD  represent  a  rectangle,  of  which  AB  is  8  units  long,  and 
AC  5.  Now,  let  us  conceive  a  square  a  constructed  on  one  of  these  units. 
Using  this  surface  as  the  unit  of  area,  it  is  evident 
that  in  the  rectangle  cABd  there  will  be  8  such. 
Hence,  the  area  of  this  rectangle  is  8  (square  units). 
Now,  drawing  parallels  to  the  base  through  the 
several  points  of  division  of  the  altitude,  it  is  evident 
that  the  whole  rectangle  ABCD  is  made  up  of  as 
many  rectangles  like  cAB^  as  there  are  units  in  the 
altitude — in  this  case  5,  Hence  the  whole  area  is  5 
times  the  area  of  cABfZ,  i.e.,  5  times  8  (square  units) 
=  40  (square  units). 

N.B. — The  pupil  should  be  careful  to  observe  that  the  language   '^product  of 
ba^e  into   cUtitude^^  is   only  a  convenient  form  of  abbreviated  expression.    It  is 


^ 


A/    ^ 


'   /f.  5    6   7   ffQ 
Fig.  82. 


48 


ELEMENTARY  GEOMETRY. 


just  as  absurd  to  talk  about  multiplying  a  line  bj^  a  line,  as  to  talk  about  multi- 
plying dollars  by  dollars.  Thus  8  inches  in  length  can  be  taken  5  times,  and 
makes  40  inches  in  length.  But  what  does  8  inches  in  length,  multiplied  by  5 
inches  in  length  mean  ?  Or  what  is  8  dollars  taken  5  dollars  times  ?  The  multi- 
plier must  always  be  an  abstract  number,  and  the  product  be  like  the  multipli- 
cand, from  the  *\'erv  nature  of  multiplication.  With  this  the  explanation  given 
above  agrees.  When  we  say  that  the  area  of  ABCD  =  8  x  5,  we  mean  5  times 
8  squai-e  units,  which  equals  40  square  units. 


100,  TJieoreni. — TJie  area  of  aiiy  parallelogram  is  the  same  as 
the  area  of  a  rectangle  having  the  same  base  and  altitude  as  the  parah 
lelogram,  and  hence  is  the  product  of  its  base  and  cdtitude. 

III. — This  truth  is   easily  illustrated  by  cutting  out  a  parallelogram,  as 

ABCD.  Then,  cutting  off  the  triangle  DEC, 
being  careful  to  make  DE  perpendicular  to 
BC,  and  placing  DC  upon  AB  so  as  to  bring 
the  triangle  DEC  into  the  position  AFB,  the 
two  parts  will  just  make  up  the  rectangle 
AFED.  Hence  we  see  that  the  area  of  ABCD 
is  the  same  as  the  area  of  AFED,  which  latter 

is  a  rectangle  having  the  same  base  AD,  and  the  same  altitude  ED,  iis  the  given 

parallelogram. 


Fig.  S3. 


107*  Tlieorenu — The  area  of  a  triangle  is  half  the  j^roduct  of 
its  base  and  altitude. 

III.— To  illustrate  this  truth,  cut  out  two  triangles  A  and  B  just  alike.    By 

placing  them  together,  a 

/P\^  /^\" " 7      parallelogram      can      be 

/  i     ^\  /  j      ^\^    A        .-'         formed    whose   base    and 

/      I    A      ^\  /      I     ^      ^\^    , '  altitude  are  the  same  as 

^ ' -^  ^ '■ the   base  and  altitude  of 

^''G.  &4.  tiie  triangle.    The  area  of 

the  parallelogram  is  the  product  of  its  base  and  altitude.     Hence  the  area  of 
one  of  the  ti'iangles  is  one-half  the  product  of  its  base  and  altitude. 

In  fact,  by  cutting  one  of  the  triangles,  as  A,  into  two  triangles,  its  parts  can 
be  put  with  B  so  as  to  make  a  rectangle  having  the  same  base  and  altitude  as 
the  ti'iangles.    [The  pupil  should  do  it.] 


108,  llieoreni,—The  area  of  a  trapezoid  is  ilie  product  of  its 
altitude  into  the  line  joiimig  the  middle  points  of  its  inclined  sides, 

III.— To  illustrate  this  truth,  cut  out  any  trapezoid,  as  ABCD,  and  through 


ABOUT  AREAS. 


49 


the  middle  of  the  inclined  sides,  as  a  and  J,  cut 
off  the  triangles  kam  and  B6/i,  being  careful  to 
cut  in  lines'  am  and  hn  perpendicular  to  the 
base.  These  can  be  applied  as  indicated  in  the 
figure,  so  as  to  fill  out  the  rectangle  omnp. 
Hence  we  see  that  the  area  of  the  trapezoid  is 


A    nv 


TV  B 


Fig.  85. 

just  equal  to  the  product  of  its  altitude  into  the  line  joining  the  middle  points 
of  its  inclined  sides,  as  ah. 

Ex.  1.  How  many  square  yards  of  plastering  in  the  walls  of  a 
room  20  feet  by  30,  and  15  feet  high,  including  the  ceiling  ? 

Ans.  233f 
Ex.  2.  A  salesman  is  selling  a  piece  of  velvet  which  is  worth  88 
per  yard.  The  velvet  is  cut  "  on  the  bias,"  as  the  technical  phrase 
is,  i.  e.,  obliquely,  instead  of  square  across.  The  piece  he  is  selling  is 
measured  along  the  selvedge  in  the  usual  way  half  a  yard.  He  is 
disposed  to  charge  the  customer  somewhat  more  than  84.  Is  he 
right  ?  The  customer  claims  that  he  is  getting  but  half  a  yard  of 
velvet,  and  so  ought  to  pay  but  84.    Is  he  right  ? 

Ans.  Both  are  right, — the  salesman  in  his  demand,  and  the 
customer  in  his  statement.     How  is  it  ? 

Ex.  3.  There  are  two  parallel  roads  one  mile  apart.  A  has  a  farm 
which  extends  along  one  of  the  roads  half  a  mile,  and  the  lines  run 
perpendicularly  from  one  road  to  the  other.  B  has  a  farm  l}ing  be- 
tween the  same  roads,  and  half  a  mile  front  on  each  road,  but  run- 
ning obliquely  across.  Which 
has  the  larger  farm  ?  C  D  E  F 

Ex.  4.  Of  the  four  triangles 
ACS,  ADB,  AEB,  and  AFB,  Fig. 
86,  which  has  the  greatest 
area,  CF  being  parallel  to  AB  ?  fig.  sg. 

Ex.  5.  Which  is  the  largest  triangle  which 
can  be  inscribed  in  a  semicircle,  having  the 
diameter  for  its  base  ? 

Ex.  6.  Can  you  vary  the  area  of  a  triangle 
while  the  sides  remain  of  the  same  length? 
Can  you  vary  the  area  of  a  quadrilateral  while  the  sides  remain  of  the 
same  length  ? 

Ex.  7.  If  you  have  two  lines  each  5  inches  long,  and  two  each  3 
inches  long ;  into  what  kind  of  a  parallelogram  must  you  form  them 
in  order  to  have  its  area  the  greatest  ? 

4 


Fig.  87 


50 


ELEMENTARY   GEOMETRY. 


Ex.  8.  Rough  boards  are  usually  narrower  at  one  end  than  at  the 
other,  for  which  reason  the  lumberman  measures  their  width  in  the 
middle.     What  is  the  number  of  square  feet  in  the  following  : 
12  boards  16  feet  long,  10  inches  wide  (in  the  middle) ; 
15  boards  11  feet  long,    9  inches  wide        "         "        ; 
8  boards  10  feet  long,  13  inches  wide        "         "        ? 
What  principle  is  involved  in  such  measurement  ? 

Ex.  9.  What  is  the  area  of  a  triangle  whose  altitude  is  6  feet,  and 
base  10  feet  ?  Are  these  elements  sufficient  to  fix  the  form  of  the 
triangle  ? 

Ex.  10.  If  a  line  be  drawn  from  any  angle  of  a  triangle  to  the 
middle  of  the  opposite  side,  what  is  the  relation  of  the  areas  of 
the  two  partial  triangles  ?    Why  ? 


THE  PYTHAGOREAN  PROPOSITION.' 

109*  Hieorem, — Tlie  square  described  on  ilie  liyiMenuse  of  a 
right  angled  triangle  is  equivalent  to  the  sum  of  tlie  tico  squares 
described  on  the  other  two  sides. 

III. — The  meaning  of  this  proposition  may  be  illustrated  thus  :  Let  ABC  be 
a  right  angled  triangle,  right  angled  at  C,  and  the 
sides  AC  and  CB  be  4  and  3  respectively.  Then 
measuring  AB,  it  will  be  found  to  be  5,  and  we 
observe  that  4"^  +  S''  =  5*.  This  is  also  seen  from 
the  figure,  in  which  the  square  on  AC  contains 
4"=  16  square  units,  and  that  on  CB  3^=  9 ;  while 
that  on  AB  contains  5'  =  25,  i.  e.,  as  many  as  on 
both  the  other  sides.  We  cannot  so  readily  illus- 
trate the  truth  of  the  proposition  when  the  ratio 
of  the  sides  is  any  other,  than  that  of  3,  4,  and  5, 
but  it  is  equally  true  in  all  cases,  as  will  be  proved 
in  the  next  part  of  this  book. 

Ex.  1.  Can  you  make  a  right  angled  triangle  whose  sides  shall  be 
5,  8,  and  10  ? 

SuG.— As  10  is  the  longest  side,  it  will  have  to  be  the  hypotenuse.  Now  5^ 
+  8*  =  25  +  64  =  89.  But  10''^  =  100.  Hence,  10  is  too  long  for  the  hypote- 
nuse of  a  right  angled  triangle  whose  other  sides  are  5  and  8. 

Ex.  2.  Can  you  make  a  right  angled  triangle  whose  sides  shall  be 
9,  12,  and  15? 


THE  PYTHAGOREAN   PROPOSITION. 


51 


Ex.  3.  A  carpenter  has  framed  the  four  sills  of  a  building  to- 
gether, and  placed  them  on  the  foundation.  He  then  wishes  to 
adjust  them  so  that  the  angles  shall  be 
right  angles.  He  places  one  end  of  his 
ten  foot  pole  ah  at  a,  6  feet  from  c ;  and, 
holding  it  in  position,  orders  his  attendants 
to  move  the  sill  AB  to  the  right.  How  far 
will  the  end  h  of  the  pole  be  from  c  when 
the  angle  B  is  a  right  angle  ? 

Ex.  4.  A  gate  is  to  be  10  feet  long  and  4  feet  high.  How  long 
must  the  brace  be  to  go  in  as  a  diagonal  and  hold  the  gate  in  the 
form  of  a  rectangle  ? 

Ex.  5.  The  angles  of  a  room  are  all  right  angles,  and  its  dimen- 
sions are  20  feet  by  30  on  the  floor,  and  15  feet  high.  AVhat  is  the 
length  of  the  longest  diagonal  extending  from  one  corner  on  the 
floor  to  the  opposite  corner  in  the  ceiling  ? 

Ans,  A  little  more  than  39  feet. 

Ex.  6.  The  numbers  3,  4,  and  5  are  much  used  by  artizans  as 
parts  of  a  right  angled  triangle.  Will  any  equi-multiples  of  them 
answer  the  same  purpose,  as  twice  them,  i.  e.,  6,  8,  and  10 ;  or  three 
times  them,  as  9,  12,  and  15,  etc.  ? 

Ex.  7.  In  an  obtuse  angled  triangle,  is  the  square  of  the  side  oppo- 
site the  obtuse  angle  greater  or  less  than  the  sum  of  the  squares  of 
the  other  two  sides  ?  How  is  it  with  the  square  of  the  side  opposite 
an  acute  angle  ? 


SuG.— In  the  right  angled  tinangle  ABC,  AC  = 

CB^  +  AB'^  In  the  obtuse  angled  triangle  C'B  is 
equal  to  CB  in  the  right  angled  triangle.  But  AC"* 
is  greater  than  AC^  hence  AC''  >  BC'^  +  AB^ 
By  'a  similar  inspection  the  other  case  may  be 
determined. 


Fig.  90. 


no*  JProh, —  To  find  a  mean  proportional  hetween  two  lines. 


Solution. — I  wish  to  find  a  mean  propor- 
tional between  the  lines  M  and  N,  ^.  e.,  a  line 
aj,  such  that 

M  :  a;  : :  a;  :  N,  whence  a;'  =  M  X  N,  and 
x=  -v/M  X  N. 

I  draw  a  line  AB  equal  to  the  sum  of  M  and 
N,  making  DB  =  M,  and  AD  =  N.    I  draw  a 


52  ELElfENTARY  GEOMETRY. 

semicircumference  on  AB,  and  at  D  erect  CD  perpendicular  to  AB.    CD  is  a;, 
the  mean  proportional  required. 

Ex.  1.  To  construct  a  square  which  shall  be  equal  in  area  to  a 
given  rectangle. 

SuG. — Draw  any  rectangle.  Then  find  a  mean  proportional  between  its 
adjacent  sides  as  described  above.  A  square  constructed  on  this  line  will  be 
equal  in  area  to  the  rectangle ;  since,  if  .r  is  the  side  of  the  square,  and  M  and  N 
are  the  adjacent  sides  of  the  rectangle,  x"^  =  M  x  N.  But  x^  is  the  area  of  the 
square,  and  M  x  N  is  the  area  of  the  rectangle. 

Ex.  2.  To  find  the  square  root  of  15  by  means  of  the  ruler  and 
compasses. 


SuG.— Since  15  =  3  x  5,  if  DB  =  3  and  AD  =  5,  Fig.  91,  x  (CD)  =  V  3  x  5 
=  V  15.  Therefore,  making  a  figure  having  DB  and  AD  of  these  lengths, 
CD  can  be  measured,  and  thus  the  square  root  of  15  obtained,  approximately,  in 
numbers. 

N.  B. — In  such  a  case  CD  represents  exactly  tJie  required  root,  although  we 
may  not  be  able  to  express  the  value  exactly  in  numbers.  In  this  case  geometry 
does  exactly  what  arithmetic  can  only  do  approximately. 

Ex.  3.  Draw  a  line  which  shall  represent,  exactly,  the  square  root 
of  5. 

SuG.— Make  DB  =  1,  and  AD  =  5. 

Ex.  4.  Draw  a  rectangle  whose  adjacent  sides  are  2  and  3,  and 
then  draw  a  square  of  the  same  area. 


111.  Theoi^em,— Tlie  areas  of  similar  triangles  are  to  each 
other  as  the  squares  of  their  homologous  sides. 

III.— The  meaning  of  this  is,  that  if  ABC  and  DEF 

K^^  are  similar,  and  any  side  of  ABC  is  2  times  as  great  as  the 

1    \.  homologous  side  of  DEF  (as  is  the  case  in  the  figure,  CB 

\        \.  being  =  2FE,  CA  to    2FD  and  AB  to  2DE)  the  area  of 

r;; '^\.  ^^^  i'^  4  times  the  area  of  DEF.     In  fact,  in  a  simple 

\    V.      \  \^^^  case  like  this,  we  can  di\nde  ABC  into  four  triangles 

\         '^-..j         N^  exactly  equal  to  DEF,  as  is  done  by  the  dotted  lines. 

A  B 

Ex.  1.  A  and  B  have  triangular  pieces  of  land, 

K  which  are  similar  to  each  other,  and  similarly 

1    N.  situated.     But  A's  front  is  to  B's  as  5  to  3  ;  bow 

I         X  much  more  land  has  A  than  B  ? 
°      p  ^92  Ans.  2|  times  as  much. 


AREAS   OF   SIMILAR   TRIANGLES. 


53 


Ex.  2.  In  order  that  one  triangle  may  be  similar  to  and  4  times  as 
great  as  another,  how  must  any  side 
of  the  first  compare  with  the  ho- 
mologous side  of  the  second  ? 

Ex.  3.  In  order  that  the  areas  of 
two  similar  triangles  may  be  to 
each  other  as  4  to  9,  what  must 
be  the  ratio  of  their  homologous 

.  ,       „  ^  Fig.  93. 

Sides  ? 


112.  Theore^n.—The  homologous  sides  of  similar  triangles  arc 
to  each  other  as  the  square  roots  of  their  areas. 

This  theorem  is  involved  in  the  theorem  that  the  areas  of  similar  triangles 
are  to  each  other  as  the  squares  of  their  homologous  sides.  It  is  illustrated  in 
the  preceding  examples. 

Ex.  Construct  a  triangle  with  one  of  its  sides  2  in  length. 
Then  construct  a  similar  triangle  1|  times  as  large.  What  must  be 
the  length  of  the  side  of  the  second  triangle  which  is  homologous 
with  the  side  2  of  the  first. 


Solution.— Let  CAB  be  the  given  triangle,  whose  side  AB  is  3.     Since  the 
second  is  to  be  li  times  as  great  as  the  first,  the  ratio  of  the  areas  is  2  :  3. 
Hence,    V'^  ■  V^ 
::  AB  (or    2)    :   x, 
the  side  of  the  re- 
quired triangle  ho- 
mologous with  side 
2  of  the  given  tri- 
angle.     Construct 
the  square  roots  of 
2  and  3,  as  ab  and 
ac  in    the    figure, 
and    then   find   a 
fourth  proportional 
to  ab,  ac,  and  AB. 

This  is  found  to  be  ^^^  ^^ 

ay.    Taking  DE  zr 

ay,  construct  on  it  a  triangle  DEF  similar  to  ABC,  and  it  will  be  U  times  as 
large. 


54 


ELEMENTABY   GEOMETRY. 


H 
Fig.  95. 


THE  AREA  OF  A  CIRCXE. 

113*  TJieorem, — The  area  of  a  circle  whose  radius  is  r,  is  ;rr'» 
i.e.,  3.1416  times  the  square  of  its  radius. 

III. — If  we  take  a  circle  whose  radius  is  r  and  circumscribe  about  it  a  square 
ABCD,  we  observe  that  the  area  of  this  square  is  4?"'.     Hence  we  see  that  the 
area  of  a  cu-cle  is  less  than  4  times  the  square  of  its  radius.     Again,  drawing  two 
diametei"s  EF  and  CH  at  right  angles  to  each  other, 
^  and  joining  their  exti'emities,  we  have  the  inscribed 

square  GEHF.  The  area  of  this  square  is  equal  to 
the  area  of  the  two  triangles  CEF  and  EHF.  But 
area  CEF  =  iCO  x  EF  =  ^r  x  2r  =  ^^ ;  and  in  like 
manner  EHF  =  ?•='.  Hence  area  GEHF  =  2;-='. 
TVe  thus  see  that  the  area  of  a  chcle  is  more  than 
two  times  the  square  of  the  radius.  The  area 
of  a  circle  is  therefore  somewhere  between  two 
and  four  times  the  square  of  its  radius.  Just  how 
many  times  7^  the  area  is,  we  do  not  propose  to  find 
in  this  place,  but  only  say  that  it  has  been  found  to 
be  3.1416  times  r'^.  TVe  must  also  remark  that  this 
is  not  exact ;  but  it  is  near  enough  for  practical  purposes.  In  fact,  nobody 
knows  exactly  how  many  times  the  square  of  the  radius  the  area  of  a  circle  is. 

Ex.  1.  If  Yoii  cut  from  a  square  the  largest  possible  circle,  show 
that  you  cut  away  a  little  less  thau  ^  of  the  square,  or  more  exactly 
.2146  of  it. 

Ex.  2.  What  is  the  area  in  acres  of  a  circle  whose  diameter  is  3 
miles?  ^W5. 4523.904. 

Ex.  3.  A  horse  is  so  tied  to  a  tree  that  he  can  graze  on  every  side 
of  it  to  a  distance  of  100  feet.  What  is  the  area  in  acres  over  which 
he  can  graze  ?  A7is.  A  little  less  than  J  of  an  acre. 

Ex.  4.  What  is  the  area  of  a  circle  whose  radius  is  1  ? 

[Remember  this  result.] 

Ex.  5.  Wliat  is  the  area  of  a  circle  whose  radius  is  2  ?  3  ?  4  ? 
How  do  these  areas  compare  with  the  area  of  a  circle  whose  radius 
isl? 


114.  TJieorem, — 77ie  areas  of  circles  are  to  each  other  as  the 
squares  of  their  radii. 

III. — This  is  readily  seen  from  the  last  theorem.     Thus  the  area  of  a  circle 
whose  radius  is  5  is  2o7r ;  and  of  one  whose  radius  is  6,  the  area  is  36;r.     Now, 


OF  POLYGONS.  55 

the  ratio  of  these  areas  25;r  :  SGtt  is  the  same  as  25  :  36,  i.  e.,  as  the  squares  of 
the  radii  of  the  two  circles. 

Ex.  1.  In  the  figure  the  radius  of  the  outer 

circle  is  twice  that  of  the  inner.     How  do  their  ^ — ] — -^ 

areas  compare  ?     How  do  the  4  parts  into  which  X        ;        X 

the  larger  circle  is  divided  compare  with  each  /  3  /^  ^\  2  \ 

other?  \       \     ^     )       \ 

Ex.  2.  The  radii  of  2  circles  are  3  and  5  re-  V'  ^^ —         'J 

spectively  ;  what  is  the  relation  of  their  areas  ?  ^--^.^^-^^ 

Ans.  9  :  25 ;  or  one  is  2 }  times  as  large  as  the  ^^^  ^ 
other. 

Ex.  3.  I  have  a  circle  whose  radius  is  5,  and  wish  to  make  another 
whose  area  is  twice  as  great ;  what  must  be  its  radius  ? 

Ans.  "v/50,  or  7.071  nearly. 

Ex.  4.  Can  we  compare  the  areas  of  circles  by  means  of  the  squares 
of  their  diameters  as  well  as  by  means  of  the  squares  of  their  radii  ? 
How  much  greater  is  the  square  of  the  diameter  of  any  circle  than 
the  square  of  the  radius? 

Ex.  5.  Two  5-inch  stovepipes  run  together  into  one  7-inch  pipe. 
Is  the  caj^acity  of  the  one  pipe  equal  to  that  of  the  two  ? 

Ex.  6.  Two  men  bought  grindstones  of  equal  thickness.  The 
stones  cost  $4  and  $9  respectively.  One  was  2  feet  in  diameter  and 
the  other  3.     What  was  the  difference  in  the  rates  paid  ? 


SECTION  VIIL 

OF  POLYGONS. 

115.  A  Polygon  is  a  portion  of  a  plane  bounded  by  straight 
lines. 

The  word  polygon  means  many-angled ;  so  that  with  strict  propriety  we 
might  limit  the  definition  to  plane  figures  with  five  or  more  sides.  This  limita- 
tion in  the  use  of  the  word  is  frequently  made. 

116 *  A  polygon  of  three  sides  is  a  triangle  ;  of  four,  a  quadrilat- 
eral; of  five,  a  pentagon  ;  of  six,  a  hexagon  ;  of  seven,  a  heptagon  ; 
of  eight,  an  octagon;  of  nine,  a  7ionagon ;  of  ten,  a  decagon;  of 
twelve,  a  dodecagon. 


56 


ELEMENTAEY  GEOMETRY. 


11'^.  A  Regular  Polygon  is  a  polygon  whose  sides  are 
equal  each  to  each,  and  whose  angles  are  equal  each  to  each. 

US,  Hie  VeriTYieter  of  a  polygon  is  the  distance  around  it, 
or  the  sum  of  the  bounding  lines. 


119,  Theorem, — Any  polygon  may  he  divided  hy  diagonals 
df'aiun  from  any  angle,  into  as  many  triangles  as  the  polygoji  lias 
sides,  less  tivo  sides. 


Iix. — In  the  figure  the  polygon  has  7  sides. 
By  drawing  the  diagonals  from  C  to  the  other 
angles,  we  divide  the  polygon  into  5  (7—2) 
tiiangles. 


Fig.  97. 


120,  Theore^n,—The  sum  of  the  an- 
gles of  any  polygon  is  twice  as  many  right 
angles  as  the  ptolygon  has  angles  {or  sides), 
lessfonr  right  angles. 


III. 


•Draw  a  polygon,  as  ABCDEFC,  and  the  arcs  a,  b,  c,  d,  e,f,  g,  measuring 
its  angles.  With  the  same  radius  draw  a 
circle.  Beginning  at  some  point,  as  O, 
lay  off  OA  =  a,  AB  =  i,  EC  =  c,  CD  =  d, 
DE  =  e,  EF  =/,  and  FG  =  g.  It  is  found 
in  this  case  that  the  sum  of  these  meas- 
ures is  two  circumferences  and  a  half. 
Now,  one  circumference  is  the  measure 
of  4  right  angles.  Hence,  2^  circumfer- 
ences measure  2^  x  4  r=  10  right  angles. 
Thus  it  appears  that  the  sum  of  all  the 
angles  of  the  polygon  is  10  right  angles. 
This  agrees  with  the  theorem ;  for,  .by 
that,  the  sum  should  be  2  right  angles  x  7 
—  4  right  angles,  which  is  10  right  angles. 


^121,  Troh. 

lar  polygon. 


To  draw  a  regu- 


Fig.  98. 


as  the  pol5'gon  has  sides. 
of  the  polygon. 


Solution. — Draw  a  circle,  and  divide 

the  circumference  into  as  many  equal  arcs 

The  chords  of  these  arcs  will  constitute  the  perimeter 


OF  POLYGONS.  67 

The  practical  difficulty  lies  in  dividing  the  circumference  as  required.  The 
circumference  can  be  divided  into  6  equal  arcs  by  (J5).  Drawing  radii  to  these 
points  of  division,  and  bisecting  the  included  angle,  a  division  into  13  equal 
parts  is  effected.  These  can  be  again  bisected,  and  the  division  into  24  equal  parts 
effected,  etc.  Again,  the  circumference  can  be  divided  into  4  equal  parts  by 
drawing  two  diameters  at  right  angles  to  each  other  (see  Fig.  95).  These  arcs 
can  be  bisected  as  indicated  above,  and  the  division  into  8  equal  parts  effected. 
Bisecting  the  latter  arcs,  we  have  16  equal  parts,  etc.  There  is  also  a  way  to 
divide  the  circumference  into  10  equal  parts,  but  it  is  too  difficult  to  be  given 
here.  For  all  regular  polygons  except  those  of  3,  6, 12,  24,  etc.,  and  4,  8, 16,  etc., 
sides,  the  pupil,  at  this  stage  of  his  progress,  is  expected  to  eflfect  the  division 
by  trial. 


EXERCISES. 

.  1.  By  drawing  diagonals  from  any  one  angle,  into  how  many  tri- 
angles can  a  pentagon*  be  divided  ?  Show  it  with  a  figure.  Into 
how  many  an  octagon  ?     A  dodecagon  ?     A  nonagon  ?     A  hexagon  ? 

2.  What  is  the  sum  of  the  angles  of  a  hexagon  ?  Determine  the 
number  mentally,  and  then  measure  the  angles  geometrically,  as  in 
the  sohition  of  (120),  observing  that  the  latter  result  verifies  the 
former.  In  like  manner  determine  the  sum  of  the  angles  of  a  pen- 
tagon. Of  an  octagon.  Of  a  decagon.  Of  a  nonagon.  Of  a  tri- 
angle.    Of  a  quadrilateral. 

3.  If  the  angles  of  a  hexagon  are  equal  each  to  each — that  is,  if 
the  hexagon  is  equiangular — what  is  the  value  of  any  one  angle  ? 

Ans.  IJ  right  angles. 
[Note. — A  regular  polygon  is  equiangular.] 

4.  What  is  the  value  of  any  angle  of  a  regular  octagon?  Of  a 
regular  pentagon  ?     Of  a  regular  dodecagon  ? 

Anstuer  to  the  last,  If  right  angles. 

5.  Construct  a  regular  dodecagon. 

6.  Construct  a  regular  heptagon. 

Sdg's.— Observing  that  as  the  chord  for  the  hexagon 
is  the  radius,  and  hence  the  chord  for  the  heptagon  is 
a  little  less,  we  can  readily  find  by  ^rm^  just  how  wide 
to  open  the  dividers  so  that  they  shall  step  around 
the  circumference  at  7  steps.  This  is  not  a  veiy 
scientific  Avay  of  constructing  a  figure,  it  is  true,  but 
it  is  the  only  way  we  can  get  the  chord  in  this  case.  f7o.  99. 

*  Polygons  are  not  to  be  assumed  regular  unless  they  are  so  designated. 


58 


ELEMENTARY  GEOMETRY. 


7.  Construct  a  regular  octagon. 
SuG.-See  the  general  solution  {121). 


Fia.  101. 


8.  Construct  a  regular  nonagon. 

Solution.— First  get  a  quarter  of  the  cir- 
cumference by  marking  the  points  where  two 
diameters  at  right  angles  to  each  other  would 
cut  the  circumference.  AX  is  an  arc  of  90°. 
Then  from  A  take  AY  =  60°  by  using  radius  as 
a  chord.  YX  is  therefore  an  arc  of  30°.  Divide 
this  into  three  equal  parts  by  trial.  Measure 
YB  equal  to  two- thirds  of  YX,  and  AB  and  BC 
are  arcs  of  40°,  and  the  chords  AB  and  BC  are 
chords  of  the  regular  nonagon. 


9.  To  draw  a  five-point  star. 

Solution.— Draw  a  circle,  and  dividing  the 
circumference  into  five  equal  parts,  join  the 
alternate  points  of  division,  as  in  the  figure. 


10.  To  circumscribe  a  square  about  a  circle  {56)-    Also  an  equi- 
lateral triangle,  and  a  regular  hexagon. 


SYNOPSIS  OF  PLANE  FIGURES. 


59 


SYNOPSIS  OF  I^LANE  FIGURES. 


^  What? 


O 

< 
Oh 


03 

o 

o 

^^ 
o 


0^ 

Q 

w 

Q 

P 
1^ 

O 

pq 


What? 


Triakgles. 


Sides.  Perimeter. 

What  ?     Altitude. 
'  Scalene. 

Isosceles. 

Equilateral. 


Diagonal. 
Base. 

^3  ^  r  Acute. 

I  t    Right. 


Quadrilat- 
erals. 


Pentagon. 
Hexagon. 
Heptagon. 
Octagon. 
Nonagon,  etc. 


What? 

Trapezium. 

Trapezoid. 


Parallelo-  J 
gram.        ' 


f  Rhombus. 
Rhomboid. 


Rectan- 


i  With   unequal 
\      sides. 


T      \      sides, 
t     g^l^^^- 1  Square. 


Regular.    What? 


Circle. 


What? 

Circumference. 

Centre. 

Radius,  Diameter. 


r  Ellipse. 
CoKic  Sections.*  \  Parabola. 
( Hyperbola. 

Higher  Plane  Curves.* 


*  These  are  inserted  simply  to  give  completeness.    Of  course,  the  student  ia  not  expected 
to  know  more  than  their  names. 


PART  II. 


THE  FUIN^DAMENTAL  PEOPOSITIONS  OF  ELEMENT- 
AEY  GEOMETRY,  DEMONSTRATED,  ILLUS- 
TRATED, AND  APPLIED. 


CHAPTER  I. 

FLAXE  GEOMETRY. 


SECTION  L 

OF  PERPENDICULAR  STRAIGHT  LINES. 


PROPOSITION  L 

122.  OTJieoreni. — At  any  point  in  a  straight  line,  one  perpen- 
dicular can  he  erected  to  the  line,  and  only  one,  luhich  shall  lie  on  the 
same  side  of  the  line. 

Dem. — Let  AB*  represent  any  line,  and  P  be 
any  point  therein ;  then,  on  the  same  side  of 
AB  there  can  be  one  and  only  one  perpendic- 
ular erected  at  P.  For  from  P  draw  any  ob- 
lique line,  as  PC,  forming  with  AB  tlie  two 
angles  CPB  and  CPA.  Now,  while  the  ex- 
Q  tremity  P,  of  PC,  remains  at  P,  conceive  the 
line  PC  to  revolve  so  as  to  increase  the  less  of 
Fig.  102.  the  two  angles,    as   CPB,  and    decrease    the 

greater,  as  CPA.     It  is  evident  that  for  a  certain  position  of  CP,  as  C'P,  these 


c 

/c 

/ 

/ 
/ 

-..   A 

r 

*  In  class  recitation  the  pnpil  shoiiM  jjo  to  the  blackboard,  after  having  had  hip  proposition 
assigned  hira,  and  first  draw  the  figure  required  for  the  demonstration.  This  ehould  be  done 
neatly,  accurately,  with  dispatch,  and  ivifhout  any  aids.  The  figure  being  complete,  he 
stands  at  the  board,  pointer  in  hand,  enunciates  the  proposition,  and  then  gives  the  demon- 
Btration  as  it  is  in  the  text,  pointing  to  the  several  parts  of  the  figure  as  they  are  referred  to. 


OF  PERPENDICULAR  LINES. 


61 


angles  will  become  eqn<al.  In  this  position  C'P  becomes  perpendicular  to  AB 
{26).*  Again,  if  the  line  C'P  revolve  from  the  position  in  which  the  angles 
are  equal,  one  angle  will  increase  and  the  other  diminish ;  hence  there  is  only 
one  position  of  the  line  on  this  side  of  AB  in  which  the  adjacent  angles 
are  equal.  Therefore  there  can  be  one  and  only  one  perpendicular  erected  to 
AB  at  P,  which  shall  lie  on  the  same  side  of  AB.    q.  e.  d. 

123,  Cor.  1. —  On  the  other  side  of  the  line  a  second  xjerpen- 
dicular,  and  only  one,  can  he  draimi  from  the  same  point  in  the  line. 

124z.  Cor.  2. — If  one  straight  line  meets  another  so  as  to  mahe 
the  angle  on  one  side  of  it  a  right  angle^  the  angle  on  the  other  side  is 
also  a  right  angle,  and  the  first  line  is  2^erpendic2ilar  to  the  second. 

12S,  Cor.  3. — If  two  lines  intersect  so  as  to  make  one  of  the 
four  angles  formed  a  right  angle,  the  other  three  are  right  angles,  and 
the  lines  are  mntuaUy  2)erpendicular  to  each  other, 

Dem.— Thus,  if  CEB  is  a  right  angle,  CEA, 
being  equal  to  it,  is  also  a  right  angle.  Then, 
as  AEC  is  a  right  angle,  the  adjacent  angle 
AED  is  a  right  angle,  since  they  are  equal. 

Also,  as  CEB  is  a  right  angle,  and  BED  equal        

to  it,  BED  is  a  right  angle.     Hence   CD  being       A 
perpendicular  to  AB,   AB   is  perpendicular  to 
CD,  as  it  meets  CD  so  as  to  make  the  adjacent 
angles  AEC  and  AED,  or  CEB  and  BED  equal  to 
each  other  (45). 

Fig.  10:j. 


C! 


PROPOSITION  II. 

126,  Tlieorem, —  Wlien  two  straight 
lines  intersect  at  right  angles,  if  the  por- 
tion of  the  2)iane  of  the  lines  on  one  side 
pf  either  line  he  conceived  as  revolved  on 
that  liiie  as  an  axis  until  it  coincides 
tvith  the  portion  of  the  plane  on  the  other 
side,  the  parts  of  the  second  line  will  coin- 
cide. 


Dem. — Let  the  two  lines  AB  and  CD  intersect 
at  right  angles  at  E  ;  and  let  the  portion  of  the 
plane  of  the  lines  on  the  side  of  CD  on  which 
B  lies  be  conceived  to  revolve  on  the  line  CD  as  an  axis,!  until  it  falls  in  the 


D 

Fig.  104. 


*  When  a  preceding  principle  is  referred  to,  it  should  be  accurately  quoted  by  the  pupil. 

+  As  if  tlie  paper,  which  may  represent  the  plane  of  the  lines,  were  folded  in  the  line  CD- 
It  is  important  that  this  process  be  clearly  conceived,  as  it  is  to  be  made  the  basis  of  many 
Bubsequent  demonstrations. 


62 


ELEMENTARY  PLANE  GEOMETRY. 


portion  of  tJ^  plane  on  the  other  side  of  CD.     Then  will  EB  fall  in  and  coincide 
with  AE. 

For,  the  point  E  being  in  CD,  does  not  change  position  in  the  revolution ; 
and,  as  EB  remains  perpendicular  to  CD,  it  must  coincide  with  EA  after  the 
revolution,  or  there  would  be  two  peipendiculars  to  CD  on  the  same  side  and 
from  the  same  point,  E,  which  is  impossible  {122).  Hence  EB  coincides  with 
EA.     Q.  E,  D. 


PROPOSITION  in. 

127,  Tlieorem, — From  cuiy  point  without  a  straiglit  line,  one 
-perpendicular  can  he  let  fall  %ipon  that  tine,  and  only  one. 

Dem. — Let  AB  be  any  line,  and  P  any  point  without  the  line  ;  then  one  per- 
pendicular, and  only  one,  can  be  let  foil  from  P 
P  upon  AB. 

For,  conceive  any  oblique  line,  as  PC,  drawn, 
making  the  angle  PCB>PCA.  Now,  while  the 
extremity  P  of  this  line  remains  fixed,  conceive 
the  line  to  revolve  so  as  to  make  the  greater  angle 
PCB  decrease,  and  the  less  angle  PC  A  increase. 
At  some  position  of  the  revolving  line,  as  PD,  the 
two  angles  which  it  makes  with  the  line  AB  will 
become  equal.  When  these  adjacent  angles  are  equal,  the  line,  as  PD,  is  per- 
pendicular to  AB  {20 f  4:3).  Moreover,  there  is  only  one  position  of  the  line  in 
which  these  angles  are  equal ;  hence,  only  one  perpendicular  can  ])e  drawn 
from  a  given  point  to  a  given  line.    q.  e.  d. 


D       E 

Fig.  105. 


C    B 


Ps 


Fig.  106. 


PROPOSITION  IT. 

128,  Theorein.—From  a  point  with- 
out  a  straight  line,  a  perpendicular  is  the 
shortest  distance  to  the  line. 

Dem. — Let  AB  be  any  straight  line,  P  any  point 
without  it,  PD  a  perpendicular,  and  PC  any 
oblique  line  ;  then  is  PD  <  PC. 

Produce  PD,  making  DP'  =  PD,  and  draw  P'C. 
Then  let  the  portion  of  the  plane  of  the  lines 
above  AB  be  revolved  upon  AB  as  an  axis,  until  it 
coincides  with  the  portion  below  AB.  Since  PP' 
and  AB  intersect  at  right  angles,  PD  will  fall  in 
DP'  {126) ;  and,  since  PD  =  DP',  P  will  fall  at  P', 
and  PC  =  P'C,  since  they  coincide  when  applied. 
Finally,  PP'  being  a  straight  line,  is  shorter  than 


OF    PERPENDICULAR    LINES. 


63 


PCP',  which  is  a  broken  line,  since  a  straight  line  is  the  shortest  distance  be- 
tween two  points.  Hence  PD,  the  half  of  PP',  is  less  than  PC,  the  half  of  the 
broken  lin^  PCP'.     q.  e.  d. 


PROPOSITION  V. 

129.  Uieorem. — If  a  perpendicular  he  erected  at  the  middle 
point  of  a  straight  line, 

1st.  Any  pfoint  in  the  perpendicular  is  equally  distant  from  the 
extremities  of  the  line. 

2(1.  Any  point  without  the  perpendicular  is  nearer  the  extremity  of 
the  line  on  its  own  side  of  the  perpendicular. 

Dem. — 1st.  Let  PD  be  a  perpendicular  to  AB  at  its  middle  point  D.  Then, 
O  being  any  point  in  the  perpendicular,  OA  =  OB. 

For,  revolve  the  figure  OBD  upon  OD  as  an  axis 
until  it  falls  in  the  plane  on  the  other  side  of  PD. 
Since  ODB  and  ODA  are  right  angles,  DB  will  fall 
in  DA  {126) ;  and,  since  DB  :=  DA,  B  will  fall  at  A. 
Hence,  OA  and  OB  coincide,  and  OA  =  OB. 

2d.  O'  being  any  point  without  the  perpendicular 
on  the  same  side  as  B,  0'B<0'A. 

For,  drawing  O'A  and  O'B,  let  O  be  the  point  at 
which  O'A  cuts  the  perpendicular.  Draw  OB.  Now 
O'B  < BO +  00',  since  O'B  is  a  straight  and  O'OB  is  a  broken  line.  But,  as 
OA=OB,  we  may  substitute  it  in  the  inequality,  and  have  O'B  <0A  +  00',  which 
sum  =  O'A. 


130.  Cor. — //  each  of  two  points  in  one  line  is  equally  distant 
from  the  extremities  of  another  line,  the  former  line  is  perpendicular 
to  the  latter  at  its  middle  point. 

'Dem.— Every  point  equally  distant  from  the  extremities  of  a  straight  line  lies 
in  a  perpendicular  to  that  line  at  its  middle  point,  by  the  proposition.  But, 
two  points  determine  the  position  of  a  straight  line.  Hence,  two  points,  each 
equally  distant  from  the  extremities  of  a  straight  line,  determine  the  position 
of  the  perpendicular  at  the  middle  point  of  the  line. 


A 


EXERCISES. 

1.  JProh. — To  erect  a  perpendicular  to  a  given  line  at  a  given 
voint  in  the  line. 


64: 


ELEMENTARY  PLANE  GEOMETRY. 


A  C 

Fig.  108. 


Solution. — [The  process  is  given  in 
{44),  and  should  be  repeated  here  ex- 
actly as  given  there,  with  the  reasons  for 
the  solution,  as  follows.]  A  is  one  point 
in  the  line  OA,  which  is  equally  distant 
from  B  and  C,  by  construction,  and  0  is 
another.  Hence  OA  is  perpendiculai*  to 
BC  at  A,  by  [130). 


2.  I* rob. — To  bisect  a  given  line. 


Fig.  109. 


Solution. — [For  the  process  see  {39).  The 
student  should  first  do  it  as  he  did  then.  The 
reason  why  this  process  bisects  AB  is  as  follows.] 
Since  m  is  one  point  equally  distant  from  the  ex- 
tremities A  and  B,  and  n  another,  there  are  two 
points  in  mn  each  equally  distant  from  the  ex- 
tremities of  AB.  Hence  mn  \s,  perpendicular  to 
AB  at  its  middle  point  0,  by  {130).  [The  reason 
for  the  process  in  Fig.  20  is  the  same.  Let  the 
student  give  this  method,  and  show  how  the  cor- 
ollary {130)  applies.] 


XT 


Fig. 110. 


Cx 


3.  I^roh^ — From  a  point  witliout  a 
given  line,  to  let  fall  a  perpendicular  upon 
the  line. 

Solution. — [Repeat  the  process  as  in  (45), 
and  give  the  reason  for  it  as  follows.]  O  is  one 
point  equally  distant  from  B  and  C,  and  D  is 
another.  Hence  a  line  drawn  from  0  to  D  is 
perpendicular  to  BC  by  {130). 


4.  Wishing  to  erect  a  line  perpendicular  to  AB  at  its  centre,  I 

take  a  cord  or  chain  somewhat 


P' 


ft^ 


P 

Fig.  111. 


longer  than  AB,  and,  fastening 
its  ends  at  A  and  B,  take  hold  of 
the  middle  of  the  cord  or  chain 
and  carry  it  as  far  from  AB  as  I 
can,  first  on  one  side  and  then  on 
the  other,  sticking  pins  at  the 
most  remote  points,  as  at  P  and 
P'.    These  points  determine  the 


perpendicular  sought.    What  is  the  principle  involved? 


5.  Two  boys  are  skating  together  on  the  ice,  and  both  start  from 


OF  OBLIQUE  LINES.  65 

the  same  point  at  the  same  time,  one  skating  directly  to  the  shore 
and  the  other  obliquely.  They  both  reach  the  shore  at  the  same 
time.     Which  skates  the  faster  ?    What  principle  is  involved  ? 

6.  Several  persons  start  at  different  times  from  the  same  point  in 
a  straight  road  that  runs  along  a  wood,  and  each  travels  directly 
away  from  the  road.  Will  they  come  out  at  the  same,  or  at  different 
points  on  the  opposite  side  of  the  wood  ?  What  principle  is  involved  ? 
What  is  the  geometrical  language  for  the  colloquial  phrase  "  Directly 
away  from  the  road"  ? 

7.  If  I  go  from  A  to  B,  Fig,  111,  by  first  passing  over  AP,  will  I 
gain  or  lose  in  distance  by  going  on  a  little  farther  in  the  direction 
of  AP  before  I  turn  and  go  straight  to  B  ?  What  principle  is  in- 
volved ?  Would  I  gain  or  lose  by  stopping  short  of  P  on  the  line 
AP?    Why? 


SECTION  II. 

OF  OBLIQUE   STRAIGHT  LINES. 


PROPOSITION  I. 

131,  Theore^n, —  When  an  oblique  line  meets  another  straight 
line  forming  tzuo  adjacent  angles,  the  sum  of  these  angles  is  two  right 
angles. 

Dem. — Let  the  oblique  line  CD  meet  the  straight  , 

line  AB  forming  the  two  adjacent  angles  CDB  and 
CDA  ;  then  CDB  +  CDA  equals  two  right  angles. 

For  suppose  CD  to  revolve  toward  the  position  of 
the  perpendicular  CD  ;  the  angle  CDB  will  increase 

at  the  same  rate  that  CDA  diminishes;  hence  their     „ 

sum  will  remain  constant  {i.  e.,  the  same).     But, 

when  CD  becomes  perpendicular,  the  sum  of  the 

adjacent  angles  formed  with  AB  is  two  right  angles  by  definitions  {26,  43). 

Therefore  CDB  +  CDA  =  two  right  angles.    Q.  e.  d. 

132.  Cor. — The  sum  of  all  the  consecutive  angles  formed  hy  any 
number  of  lines  meeting  a  given  line  on  the  same  side  and  at  a  given 
point  is  ttvo  right  angles. 

5 


66 


ELEMENTARY  PLANE  GEOMETRY. 


D 
Fig.  113. 


Dem.— Thus  ADC"  +  C^DC'"  +  C'^DC"  +  C"'DC" 
+  CDC  +  CDC  +  CDB  =  ADC  +  CDB,  which 
sum  is  two  right  angles  by  the  proposition.  Or,  in 
general  tenns,  the  angles  thus  formed  can  always  be 
united  into  two  groups,  constituting  respectively  the 
two  adjacent  angles  formed  by  one  line  meeting 
another. 


133,  Def. — Two  angles  whose  sum  is  two  right  angles,  are 
called  StippUmenial  Angles.  Hence,  the  Supplement  of  an  angle  is 
what  remains  after  subtracting  it  from  two  right  angles. 


PROPOSITION  n. 

1.34:.  Tlieorem, —  WJien  any  tivo  straight  lines  intersect,  the 
opposite  or  vertical  angles  are  equal  to  each  other,  and  the  sum  of  thv 
four  angles  formed  is  four  right  angles. 

Dem.— Let  AB  and  CE  intersect  at  D ;  then  CDA  =  the  opposite  angle  BDE, 
ADE  =  the  opposite  or  vertical  angle  CDB,  and  ADC  +  CDB  +  BDE  +  EDA  rl 
four  right  angles.  For,  since  CD  meets  AB,  ADC  is  the  supplement  of  CDB 
{131,  138).  Also,  since  BD  meets  CE,  BDE  is  the 
supplement  of  CDB.  Hence  ADC  —  BDE.  In  a  sim- 
ilar manner  ADE  can  be  proved  equal  to  CDB.  [The 
student  should  give  the  proof.] 

Again,  since  ADC  +  CDB  =  two  right  angles,  and 
BDE  +  EDA  =  two  right  angles,  by  adding  the  cor- 
responding members  together,  we  have  ADC  +  CDB 
4-  BDE  +  EDA  =  four  right  angles. 


Fig.  114. 


135.  Cor.— The  sion  of  all  the  consecutive  angles  formed  hy  any 
number  of  lines  meeting  at  a  common  point  is  four  right  angles. 

Dem.— The  truth  of  this  corollary  is  rendered 
apparent  by  drawing  a  line  through  the  common 
vertex,  and  observing  that  the  sum  of  all  the  angles  on 
each  side  thereof  is  two  right  angles;  whence  the 
sum  of  all  the  angles  on  both  sides,  which  is  the 
same  as  the  sum  of  all  the  consecutive  angles  foi-med 
by  the  Une,  is  four  right  angles.  [Let  the  student  put 
letters  on  the  figme,and  demonstrate  by  means  of  it] 


Fig.  115. 


OF   OBLIQUE   LINES. 


67 


PROPOSITION  III. 

136.  Theorem, — If  two  supplemental  angles  are  so  situated  as 
to  le  adjacent  to  each  other,  the  two  sides  not  common  tuillfaU  in  the 
same  straight  line, 

Dem. — Let  the  sum  of  the  two  angles 
BOA  and  CO'D  be  two  right  angles. 
Prolong  CO',  forming  the  angle  DO'E. 
Then  is  DO'E  supplemental  to  CO'D  {131, 
133) f  and  hence  equal  to  BOA,  which  is 
supplemental  to  CO'D  by  hypothesis. 
Now,  if  AOB  be  placed  adjacent  to  CO'D, 
the  vertex  0  being  at  0',  and  the  side  OA 
falling  in  O'D,  OB  will  fall  in  O'E,  since 
BOA  =  DO'E.  Hence,  when  the  angles 
are  so  situated,  OB  becomes  the  prolonga- 
tion of  CO'.     Q.  E.  D. 


PROPOSITION  IV. 

137.  TJieorem. — If  from  a  point  withotit  a  straight  line  a  per- 
pendicular he  draivn,  oblique  lines  from  the  same  point  cutting  the 
line  at  equal  distances  from  the  foot  of  the  perpendicular  are  equal  to 
each  other ;  the  angles  which  they  form  iDxth  the  perpendicular  are 
equal  to  each  other  ;  and  the  angles  lohich  they  form  with  the  line  are 
equal  to  each  other, 

Dem. — Let  AB  be  any  straight  line,  P  any  point  without  it,  PD  a  perpen- 
dicular, and  PC  and  PE  oblique  lines  cutting 
AB  at  C  and  E,  so  that  DC=DE ;  then  PC  =  PE, 
angle  CPD  =  angle  DPE,  and  angle  PCD  = 
angle  PED. 

Revolve  the  figure  PDE  upon  PD  as  an 
axis,  until  it  falls  in  the  plane  on  the  other 
sic^e  of  PD.  Since  AB  is  perpehtlicular  to  PD, 
DB  will  fall  in  DA;  and,  since  DE  =  DC,  E 
will  fall  at  C.  Now,  as  P  remains  stationary, 
the  triangles  PDE  and  PDC  coincide.  Hence, 
PC  =  PE,  angle  CPD  =  angle  DPE,  and 
Angle  PCD  =  angle  PED.    q.  e.  d. 


Fig.  117. 


Query. -How  does  the  equality  of  PE  and  PC  follow  from  {129)  ? 


PROPOSITION  V. 

138.  Tlieorem.— If  from  a  point  without  a  line  a  perpendicu- 
Jar  he  draion  to  the  line,  and  also  from  the  same  point  two  ohlique 


68 


ELEMENTARY  PLANE  GEOMETRY. 


lines  making  equal  angles  with  the  perpendicular,  the  oblique  lines 
are  equal  to  each  other,  cut  the  line  at  equal  distances  from  the  foot 
<of  the  perpendicular,  and  make  equal  angles  with  it.* 

Dem.  — PD  being  a  perpendicular  to  AB,   and  angle  CPD  equal  to  angle 

DPE,  PC  equals  PE,  CD  equals  DE,  and  PCD 
equals  PED. 

Revolve  the  figure  PDE  upon  PD  as  an 

axis,  till  it  falls  in  the  plane  of  PDC.     Since 

angle  EPD  =  angle  CPD,  PE  will  take  the 

direction  PC,  and  E  will   fall  somewhere  in 

the  indefinite  line  PF.    But,  since   PDE  and 

PDC  are  right  angles,  DE  will  fall  in  DA  {126) 

and  E  will  fall  somewhere  in  the  indefinite 

line  DA.    Now,  as  E  falls  at  the  same  time  in 

^^®-  ^^^'  PF  and  DA,  it  must  fall  at  their  intersection 

C.    Hence,  PE  coincides  with  PC,  and  DE  with  DC.    Therefore  PE  =  PC,  DE 

=  DC,  and  angle  PED  —  PCD.    q.  e.  d. 


139.  TJieorem, 


PROPOSITION  VI. 

If  from  a  imnt  luithout  a  line  a  perpendicular 
he  let  fall  on  the  line,  and  tiuo  oblique 
lines  be  drawn,  the  oblique  line  which  cuts 
off  the  greater  distance  from  the  foot  of 
the  perpendicular  is  the  greater. 


Dem. — Let  AB  be  any  straight  line,  P  any  point 
without  it,  and  PC  and  PF  two  oblique  lines  of 
which  PF  cuts  off  the  greater  distance  from  the 
'^p'  foot  of  the  perpendicular;   that  is  DF  >  DC. 

Fig.  119.  Then  is  PF  >  PC. 

Produce  PD,  making  DP'  =  PD,  and  draw  FT  and  P'C,  producing  the  latter 
until  it  meets  PF  in  H.  Revolve  the  figure  FPD  upon  AB  as  an  axis,  until  it 
falls  in  the  plane  on  the  opposite  side  of  AB.  Since  PP'  is  perpendicular  to  AB, 
PD  will  fall  in  DP'  ;  and,  since  PD  =  DP',  P  will  fall  at  P'.  Then  P'C  =  PC 
and  P'F  =  PF.  Now  the  broken  line  PCP'  <  than  the  broken  line  PHP',  since 
the  straight  line  PC  <  the  broken  line  PHC.  For  a  like  reason  the  broken  line 
PHP'<PFP,  sinc^  HP' <  HFP.  Hence  PCP' <  PFP',  and  PC  the  half  of 
PCP' <  PF  the  half  of  PFP'.    Q.  E.  d. 

ScH. — If  the  two  oblique  lines  to  be  compared  lie  on  diflTerent  sides  of  the 
perpendicular,  as  PF  and  PE,  DF  being  greater  than  DE,  lay  ofi*  DC  =  DE,  and 
draw  PC.  Then  since  PC  =  PE,  if  it  is  found  less  than  PF,  as  in  the  demon- 
stration, PE  is  less  than  PF. 

*  This  proposition  is  the  converse  of  the  last.  The  eigniflcance  of  this  statement  will  be 
more  fully  developed  farther  on  ( 15-i). 


OF  OBLIQUE   LINES.  69 

140,  Cor.  1. — From  a  given  point  without  a  line,  there  can  7ioi 
he  tioo  equal  oblique  lines  draiun  to  the  line  on  the  same  side  of  a  per- 
pendicular from  the  point  to  the  line. 

14: !•  Cor.  2. — Tv)0  equal  oblique  lines  drawn  from  the  same 
point  in  a  perpendicular  to  a  given  line,  cut  off  equal  distances 
on  that  line  from  the  foot  of  the  perpendicular, 

Dem.— For,  if  the  distances  cut  oflF  were  unequal,  the  lines  would  be  unequal. 


EXERCISES. 

1.  Having  an  angle  given,  how  can  you  construct  its  supplement  ? 
Draw  any  angle  on  the  blackboard,  and  then  construct  its  supple- 
ment. 


Fig.  120. 

2.  The  several  angles  in  the  figure  are  such  parts  of  a  rignt  angle 
as  are  indicated  by  the  fractions  placed  in  them.  If  these  angles 
are  added  together  by  bringing  the  vertices  together  and  causing 
the  adjacent  sides  of  the  angles  to  coincide,  how  will  MA  and  CN 
lie  ?  Construct  seven  consecutive  angles  of  these  several  magni- 
tudes.    How  do  the  two  sides  not  common  lie  ?    Why  ? 

3.  If  two  times  A,  B,  two  times  D,  three  times  E,  three  times  C,  three 
times^C,  two  times  F,  in  the  last  figure,  are  added  in  order,  how  will 
AM  and  CN  lie  with  reference  to  each  other  ?     Why  ? 

Ans,  They  will  coincide. 

4.  If  you  place  the  vertices  of  any  two  equal  angles  together  so  • 
that  two  of  the  sides  shall  extend  in  opposite  directions  and  form 
one  and  the  same  straight  line,  the  other  two  sides  lying  on  opposite 
sides  thereof,  how  will  the  latter  sides  lie  ?     By  what  principle  ? 

5.  Upon  what  principle  in  this  section  may  the  common  method 
of  erecting  a  perpendicular  at  the  middle  of  a  straight  line  (30 ^  44) 
be  erplained  ?  Upon  what  the  method  of  letting  fall  a  perpendicular 
upon  a  straight  line  from  a  point  without  (45)  ? 


70  ELEMENTARY  PLANE  GEOMETRY. 

6.  A  and  B  start  at  the  same  time,  from  the  same  point  in  a  certain 
road  ;  A  travels  directly  to  a  point  in  another  road  at  right  angles  to 
the  first,  and  at  ten  miles  from  their  intersection,  and  B  travels  di- 
rectly toward  a  second  point  in  the  second  road,  which  point  is  seven 
miles  from  the  intersection.  Both  reach  their  destination  at  the 
same  time.    Which  travels  the  faster  ?     What  principle  is  involved  ? 


SECTION  III. 

OF   PARALLELS. 


G 


H 


PROPOSITION  I. 

14:2,  Tlieorem* — Tiuo  straight  lines  lying  in  the  same  plane 
and  peri^endiciilar  to  a  third  line  are  parallel  to  each  other. 

Dem. — Let  AB  and  CD  be  two  straight  lines 
^  \y\ng  in  the  same  plane  and  each  perpendicu- 

lar  to  FE ;  then  are  they  parallel. 

For  if  AB   and  CD  are  not  parallel,  they 
will   meet  at  some  point  if  sufficiently  pro- 

g      duced  {60).    But,  if  they  could  meet,  we  should 

^  have  two  straight  lines  from  one  point  (their 

YiG.  121.  point  of  meeting),  perpendicular  to  the  same 

straight  line,  which  is  impossible  (1;27).  There- 
fore, as  the  lines  lie  in  the  same  plane  and  cannot  meet  how  far  soever  they  be 
produced,  they  are  parallel,     q.  e.  d. 

143,  Cor.  1. — Throxigh  the  same  point  one  parallel  can  always 
be  drawn  to  a  given  line,  and  only  one, 

Dem. — Let  AB  be  the  given  line,  and  C  the  given  point,  there  can  be  one 
and  onl}'-  one  perpendicular  through  G  to  AB  {127)  Let  this  be  FE.  Now 
through  C  one  and  only  one  perpendicular  can  be  drawn  to  FE.  Let  this  be 
CD.  Then  is  CD  parallel  to  AB  by  the  proposition.  That  there  is  only  one 
Buch  parallel,  we  shall  assume  as  axiomatic* 

144.  Cor.  2. — If  a  straight  line  is  perpendicular  to  one  of  two 
parallels,  it  is  perpendicular  to  the  other  also. 

Dem. — If  FE  is  perpendicular  to  AB  it  is  perpendicular  to  CD.  For,  if 
through  G  where  FE  intersects  CD,  a  perpendicular  be  drawn  to  FE,  it  is  par- 

*  Nous  regarderons  cette   proposition   comme    evidexte.     P.-F.  Compagnon.     So  also 

CaAUVENET. 


OF  PARALLEL  LINES. 


71 


allel  to  AB  by  the  proposition.  But,  by  Cor.  1,  there  can  be  but  one  line 
thi-ough  C.  parallel  to  AB.  Hence  the  perpendicular  to  FE  at  C  coincides  with, 
oris,  the  parallel  CD. 


PROPOSITION  II. 

j[4eT.  TTieorein, — T^oo  straight  lines  ivliicli  are  parallel  to  a 
third,  are  parallel  to  each  other. 

Dem.— Let  AB  and  CD  be  each  parallel  to  EF  ; 
then  are  they  parallel  to  each  other. 

For  draw  HI  perpendicular  to  EF  ;  then  will  it 

be  perpendicular  to  CD  because  CD  is  parallel  to 

EF.    For  a  like  reason  HI  is  perpendicular  to  AB. 

Hence  CD  and  AB  are  both  perpendicular  to  HI, 

and  consequently  parallel,    q.  e.  d. 

Fig.  122. 


H 

E 

K 

F 

C 

L 

D 

A 

M 

1 

B 

146.  DEFINITION'S. — When  two  lines  are  cut  by  a  third  line 
the  angles  formed  are  named  as  follows : 

Exteriot*  Angles  are  those  without  the  two 
lines,  as  1,  2,  7,  and  8. 

Interior  Angles  are  those  within  the  two 
lines,  as  3,  4,  5,  and  6. 

Alternate  Exterior  Angles  are  those 
without  the  two  lines  and  on  different  sides  of  the 
secant  line,  but  not  adjacent,  as  2  and  7,  1  and  8. 

Alternate  Interior  Angles  are  those 
within  the  two  lines  and  on  different  sides  of 
the  secant  line  but  not  adjacent,  as  3  and  6,  4  and  5. 

Corresj^onding  Angles  are  one  without  and  one  within  the 
two  lines,  and  on  the  same  side  of  the  secant  line  but  not  adjacent, 
as  2  and  6,  4  and  8,  1  and  5,  3  and  7. 


Fig.  123. 


PROPOSITION  in. 

147,  TJieorem* — //  two  lines  are  cut  hij  a  third  line,  malcing 
the  sum  of  the  interior  angles  on  the  same  side  of  the  secant  line 
equal  to  two  right  angles,  the  two  lines  are  parallel. 

Dem.— Let  AB  and  CD  be  met  by  the  line  EF,  making  ECD  +  FKB  =  two 
right  angles;  then  are  A B  and  CD  parallel. 


72 


ele3o:ntaey  plane  geometey. 


H     G/ 


T 


B 


Fig.  124. 
PK  falls  in  PC-    Since  PK  =  PC,  K  will  fall  at  C 


For,  through  P,  the  middle  of  CK,  draw  HI 
perpendicular  to  AB.  Since  H  PC  and  KP I  are 
vertical  angles,  they  are  equal  b}-  {134}.  Also, 
since  CKB  and  CCK  are  both  supplements  of 
DCK,  the  former  by  hypothesis,  and  the  latter 
by  {133),  CKB  =  CCK.  Now,  conceive  the 
portion  of  the  figure  below  P,  while  remaining 
in  the  same  plane  (the  plane  of  the  paper),  to 
revolve  upon  P  (as  a  pivot)  from  right  to  left  till 
Agam,  since  KPI  =  GPH 
PI  will  take  the  direction  PH,  and  I  will  fall  in  PH,  or  PH  produced;  and,  since 
PKI  =  PCH,  Kl  will  take  the  direction  CH,  and  I  will  fall  somewhere  in  CC. 
Hence,  as  I  falls  in  both  PH  and  CC,  it  must  fall  at  their  intersection  H ;  and 
KIP  coincides  with,  and  is  equal  to  PHC.  But  KIP  is  a  right  angle  by  construc- 
tion; hence  CHP  is  a  right  angle.  Therefore,  AB  and  CD  are  both  perpendic- 
ular to  HI,  and  consequently  parallel  by  {142).    q.  e.  d. 

148,  CoE.  1. — If  two  lines  are  cut  ly  a  tliird,  making  the  sum  of 
the  two  exterior  angles  on  the  same  side  of  the  secant  line  equal  to  two 
right  angles,  the  two  lines  are  parallel 

Dem.— For,  if  FGD  +  EKB  =  two  right  angles,  EKB  is  the  supplement  of 
FGD;  so  also  is  KGD  {131,133).  Hence  EKB  =  KGD.  Again,  for  like 
reasons,  FGD  and  GKB  are  both  supplements  of  EKB  and  therefore  equal. 
Hence,  when  FGD  +  EKB  =  two  right  angles,  GKB  +  KGD  =  two  right  angles, 
and  the  lines  are  parallel  by  the  proposition.  The  same  is  true  for  FGC  and 
AKE.     [Let  the  student  prove  it.] 

149.  Cor.  2.—//*  tico  lines  are  cut  hy  a  third,  making  either 
two  alternate  interior,  or  either  two  alternate  exterior,  or  either  two 
corresjoonding  angles,  equal  to  each  other,  the  lines  are  parallel. 

Dem.— If  CCK  =  CKB,  KCD  +  CKB  =  two  right  angles,  since  CCK  +  KGD 
=  two  rio-ht  ano-lcs.  Hence  the  lines  are  parallel  by  the  proposition.  So  also  if 
KCD  =  AKC,  or  FCD  =  AKE,  or  CCF  =  EKB,  or  FCD  =  CKB,  or  CCF  rr 
AKC  the  two  Imes  are  parallel.     [Let  the  student  show  the  ti-uth  in  each  case.] 


FIG.  125.  Fio.  126. 

»  The  accompanying  fisares  will  aid  the  student  in  getting  this  conception  Fi^.  1^ 
represents  the  position  of  the  lines  after  the  revolution  has  gone  ahoat  half  a  right  angle,  and 
Fig.  126  when  the  revolution  is  almost  completed. 


OF  PARALLEL  LINES.  73 


PROPOSITION  IV. 

160.  TJie07^em. — If  two  parallel  lines  are  cut  ly  a  third  line, 
tJie  sum  of  the  interior  angles  on  the  same  side  of  the  secant  line  is 
equal  to  tivo  right  angles, 

Dem.— Let  the  parallels  AB  and  CD  be  cut  by  EF,  then  isDCK+  CKB=r  two 
right  angles. 

For,  if  DC K  is  not  the  supplement  of  CKB, 
let  LM  be  drawn  through  G  so  as  to  make        l 
MGK  that  supplement.     Then,  by  the  preced-         '"■"--- 

ing  proposition,  LM  is  parallel  to  AB  ;  and  we         C 

have  two  parallels  to  AB  through  the  point  C, 

w^hich  is  impossible  {14=3).    Hence,  as  no  line  /  

but  a  parallel  can  make  this  interior  angle  the        ^         /K 
supplement  of  the  other,  the  parallel  makes  it  E/ 

so.       Q.    E.    D.  !:<         n« 

^  ,  .  .  Fig.  127. 

[Let  the  student  demonstrate  this  proposi- 
tion as  the  preceding  was  demonstrated.     In  this  case  CD  and  AB  are  parallel 
by  hypothesis,  and  HI  being  drawn  perpendicular  to  one  is  perpendicular  to  the 
other  also.    When  K  falls  at  G,  Kl  falls  on  CG,  since  from  a  point  without  a 
line  only  one  perpendicular  can  be  drawn  to  that  line.  ] 

161.  Cor.  1. — If  t'wo  parallel  lines  are  cut  hy  a  third  line^  the 
sum  of  either  two  exterior  angles  on  the  same  side  of  the  secant  line  is 
equal  to  two  right  angles, 

Dem.— FGD  +  EKB  =  two  right  angles.  For  FGD  and  GKB  are  both  sup- 
plements of  DGK  {lS3f  130),  and  therefore  equal  to  each  other.  For  like 
reasons,  EKB  =  K6D.  Therefore,  FGD  +  EKB  =  GKB  +  DGK  -  two  right 
angles,  by  the  proposition. 

152.  Cor.  2. — If  two  parallel  lines  are  cut  hy  a  third  line,  either 
two  alternate  interior,  or  either  tivo  alternate  exterior,  or  either  tico 
coHespo7iding  angles,  are  equal  to  each  other. 

Dem.— If  CD  and  AB  are  parallel,  CGK  =  GKB.  For  each  is  the  supplement 
of  KGD,  the  former  by  (2.51),  the  latter  by  {150).  [Let  the  student  show  in 
like  manner  that  AKG  =  KGD,  FGD  =  AKE,  CGF  =  EKB,  FGD  =  GKB,  and 
CGF  =  AKG.] 

153.  Cor.  3. — Of  the  eight  angles  formed  when  one  line  cuts  two 
parallels,  the  four  acute  angles  are  equal  each  to  each,  and  the  four 
obtuse  angles;  or,  in  case  any  one  angle  is  a  right  angle,  all  the 
others  are  right  angles. 


74  ELEMENTASY  PLANE  GEOMETRY. 

154,  ScH.— The  last  two  propositions  and  their  corollaries  are  the  canwrse 
of  each  other;  i.  e.,  the  hypotheses  or  data  and  the  conclusions  or  things  proved 
are  exchanged.  Thus,  in  Prop.  III.,  the  hypothesis  is,  that  Tlie  sum  of  the  two 
interior  angles  on  the  same  side  of  the  secant  line  is  equal  to  two  right  angles  ;  and 
the  conclusion  is,  that  T7ie  two  lines  are  parallel.  Now,  in  Prop.  IV.,  the  hypoth- 
esis is,  that  Tlie  two  lines  are  parallel;  and  the  conclusion  is,  that  Tlie  sum  of  the 
two  interior  angles  on  the  same  side  of  the  secant  line  is  two  right  angles.*  [A  clear 
conception  of  this  scholium  will  save  the  student  from  confoundmg  these  i^rop- 
ositions.] 


PROPOSITION  y. 

ISo,  Tlieoreni. — If  tiuo  straight  lines  are  cut  by  a  third  IiTie 

making  the  sum  of  the  interior  angles  on  one  side  of  tlie  seca^it  line 
less  than  tivo  right  angles,  the  tivo  lines  luill  meet  on  this  side  of  the 
secant  line,  if  sufficiently  produced. 

Dem.— Let  AB  and  LM  be  cut  by  EF  making 

.p  MGK  4-  FKB  <  two  right  angles;   then  will 

/  AB  and  LM  meet  on  the  side  of  EF  on  which 

C        '""---G/ ^       MGK  and  FKB  lie,  if  sufficiently  produced. 

-.,,  For   the    angle    which    a    parallel    to  AB 

^     through  C  makes  with  EF  is  the  supplement 

^^  — B       of  FKB.     But  by  hypothesis  MGK  is  less  than 

this  supplement.    Hence  the  portion  CM,  of 

the  line  LM,  lies  within  CD,  and  will  meet 

KB  if  sufficiently  produced.    Q.  e.  d. 


./ 


Fig.  128. 


PROPOSITION  TI. 

loG»  Jlieoreni, — Two  parallels  are  everywhere  equally  distant 
from  each  other. 

Dem.— Let  E  and  F  be  any  two  points  in  the  line  CD,  and  EC  and  FH  per- 
pendiculars measuring  the  distances  between  the  parallels  CD  and  AB  at  these 
points ;  then  is  EC  =  FH. 

For,  let  P  be  the  middle  point  between  E  and  F,  and  PC  a  perpendicular  at 

*  The  learner  may  think  that,  if  a  proposition  is  true,  its  converse  is  necessarily  true  ;  and 
hence,  that  when  a  proposition  has  been  proved,  its  converse  may  be  assumed  as  also  proved. 
Now  this  is  by  no  means  always  the  case.  Although  in  a  great  variety  of  mathematical  prop- 
ositions, it  happens  that  the  proposition  and  its  converse  are  both  true,  we  never  assume  one 
from  having  proved  the  other ;  and  we  shall  occasionally  find  a  proposition  whose  converse  is 
not  true. 


OF   PARALLEL  LINES. 


75 


tliis  point.    Revolve  the  portion  of  tlie  figure  on  the  right  of  PO,  upon  PO  as 
an  axis,  until  it  falls  upon  the  plane  of  the  c  o  c- 

paper  at  the  left.  Then,  since  FPO  and  EPO      c 

are  right  angles,  PD  will  fall  in  PC ;  and,  as 

PF  =  PE,  F  will  fall  on  E.    As  F  and  E  are         

right  angles,  FH  will  take  the  direction  EC, 
and  H  will  lie  in  EC  or  EC  produced.    Also, 


H 


B 


G         0 

Fio.  129. 

as  POH  and  POG  are  right  angles,  OB  will  fall  in  OA,  and  H  falling  at  the  same 
time  in  EC  and  OA  is  at  their  intersection  G.  Hence  FH  coincides  with  and  is 
equal  to  EC.    q.  e.  d. 


EXERCISES. 

1.  I*roh» — Through  a  given  ])oiiit  to  draio  a  line  ^parallel  to  a 
given  line,  hy  the  2)rinci2)le  contained  in  Prop.  I.  of  this  section. 

Sug's. — Draw  a  straight  line  on  the  blackboard.  Designate  with  a  dot  some 
point  without  the  line.  To  draw  a  line  through  the  designated  point  and  par- 
allel to  the  given  line,  is  the  problem.  Let  fall  a  perpendicular  upon  the  line 
from  the  point.  Then  through  the  given  point  draw  a  line  perpendicular  to 
this  perpendicular.  The  latter  line  will  be  parallel  to  the  given  line.  (By  what 
proposition  ?) 

2.  I* rob. — Through  a  given  point  to  draio  a  2^ct,rallel  to  a  given 
line  hy  Prop.  III. 

Sug's. — Through  the  given  point  draw  an  oblique  line  cutting  the  given  line. 
Then  draw  a  line  through  the  given  point  making  an  angle  with  the  oblique 
line  equal  to  the  supplement  of  the  angle  which  is  included  between  the  oblique 
line  and  the  given  line,  and  on  the  same  side  of  the  former.  [Of  course  the 
student  will  be  required  to  do  the  work  on  the  blackboard,  guessing  at  nothing.^ 

3.  Proh, — Through  a  given 
point  to  draio  a  line  parallel  to  a 
given  lin^,  upon  the  pri7iciple  that 
thor  alternate  angles  made  hy  a 
secant  line  are  equal  {152). 

4.  A  bevel  is  an  instrument  much 
used  by  carpenters,  and  consists  of 
a  main  limb  AB,  in  which  a  tonsfue 


CD  is  placed,  so  as  to  open  and  shut 

like  the   blade   of  a   knife.     This 

tongue  turns  on  the  pivot  O,  which 

is  a  screw,  and  can  be  tightened  so 

as   to  hold   tlie  tongue   firmly  at  ^'" 

any  angle  with  the  limb.    The  tongue  can  also  be  adjusted  so  as 


76  ELEMENTARY  PLANE  GEOMETRY. 

to  allow  a  greater  or  less  portion  to  extend  on  a  given  side,  as  CB,  of 
the  limb.  Now,  suppose  the  tongue  fixed  in  position,  as  represented  in 
the  figure,  and  the  side  in  of  the  limb  to  be  placed  against  the 
straight  edge  of  a  board,  and  slid  up  and  down,  while  lines  are  drawn 
along  the  side  n  of  the  tongue.  What  will  be  the  relative  position  of 
these  lines  ?  Upon  what  proposition  does  their  relative  position 
depend  ?  How  can  the  carpenter  adjust  the  bevel  to  a  right  angle 
upon  the  principle  in  Peop.  I.,  Sec.  1  ?  At  what  angle  is  the  bevel 
set,  when,  drawing  two  lines  from  the  same  point  in  the  edge  of  the 
board,  one  with  one  edge  m  of  the  bevel  against  the  edge  of  the 
board,  and  the  other  with  the  other  edge  m',  these  lines  are  at  right 
angles  to  each  other  ? 

5.  Are  the  two  walls  of  a  building  which  are  carried  up  by  the 
plumb  line  exactly  parallel  ?     Why  ? 

6.  Pass  a  circumference  through  three  given  points,  as  in  {58), 
and  show  from  principles  contained  in  one  of  the  preceding  sections, 
that  0  is  equally  distant  from  A,  B,  and  C  ;  and  hence  that,  if  a  cir- 
cumference be  drawn  from  o  as  a  centre  with  a  radius  OA,  it  will 
pass  through  A,  B,  and  C. 

7.  Construct  two  triangles  of  unequal  sizes,  but  having  the  sides 
of  the  one  respectively  parallel  to  the  sides  of  the  other.  Are  they 
shaped  alike  ? 

8.  Construct  two  triangles  of  unequal  sizes,  but  having  the  sides 
of  the  one  respectively  perpendicular  to  the  sides  of  the  other.  Are 
they  shaped  alike  ? 

9.  Construct  a  parallelogram,  two  of  whose  sides  are  6  and  10. 
Can  you  construct  different-shaped  figures  with  the  same  sides  ? 


SYNOPSIS   OF  THE   THREE    PRECEDING    SECTIONS. 


77 


Perpendic- 
ulars. 


Prop. 
Prop. 
Prop. 


Prop.    V.  Point  in. 


w 

o 

M 

H 

O 
02 

O 


O 

M 

H 


SYJfOPSIS  OF  THE  THREE  PRECEDING  SECTIONS. 

Definition  (43). 

Prop.  I.  One  and  only  one  (  C(/r.  1.  Second  perp. 

to  a  given  line  at  -j  Cor.  2.  If  one  angle  is  rio-ht 
a  given  point.  (  Cor.  3.  One  of  4  angles  rijht 
II.  Revolved  perpendicular. 

III.  From  a  point  without  a  line. 

IV.  Shortest  distance  from  a  ix)int  to  a  line. 

iCor.  Two  points  equal- 
ly  distant  from  ex, 
iremities  of  a  line. 
I  Prob.  To  erect  a  perpendicular. 
I  Prob.  To  bisect  a  ifne. 
I  Prob.  To  let  full  a  perpendicular. 
L  t  Other  exercises. 

'  Prop.  I.  Sum  of  adja-   (  ^^^-  ^"^"    ^f.  ,consec.  angles 

cent  angles.       )  ^  ^  2"  ""i^  ^''^"^  ""^  ^"'^^ 
^  (  Bef.  Supplement. 

Prop.    II.  Qpp.  angles  equal.  \  ^''''^'    ^^^les    about    a 
'^^       ^        ^         I  point. 

Prop.  III.  Supplemental  angles  made  adjacent. 

Prop.  IV.  Cutting  equal  distances  from  loot  of  perpen- 
dicular. 

Prop.    V.  Making  equal  angles  with  perpendicular. 

_,  ,  r  Cor.  1.  Not  two  equal 

Prop.  VI.  Cutting  unequal  dis- 
tances from  the  foot 
of  perpendicular. 


Exercises. 


Oblique 
Lines. 


Exercises. 
f  Definition  {66) 

Prop 


Cor.  2. 


on   same  side 
of  perpendic. 
Two  equal  ob- 
lique lines. 


Parallels. 


I. 


Two  perpendiculars 
to  a  line. 


Prop, 

r 


1.  One     parallel 
through  a  point. 

2.  A  perp.  to  one 
of  two  parallels. 


II.  Two  lines  parallel  to  a  third 

f  Exterior,  Interior,  Alter- 

Defs  of  angles  formed,  i        f ^^^  ,  Exterior,    Al- 

^  I         ternatc  Interior,  Cor- 

l        responding. 

f  Cor.  1.  Sum  of  twoEx- 

I  teiior  anjjks,  two 

Prop.  III.  Sum  of  Inter.  |  right  an y^les. 

angles,     two  ■{   Cor.  2.  Two  Ah.  Inter., 
right  angles.  |  Alt.      Exter.,    or 

Correspond'g  an- 
gles equal. 
Cor.  1.  Converse  of 
Cor.  1.,  Prop.  III. 
Cor.  2.  Converse  of 
Cor.  2.,  Prop.  III. 
Cor.  3.   Of  the  eight 

angles. 
Sch.  Moaning  of  Con- 
verse. 


Prop.  IV.  Converse  of  III. 


Prop.  V.  Sum  of  Inter,  angles  <  2  right  angles. 
Prop.  VI.  Everj^where  equidistant. 
L  Exercises.— Pw6«.  1,  2,  3.    Methods  of  drawing. 


78 


ELEMENTARY  PLANE   GEOilETRY. 


SECTION   IV. 

OF  THE  RELATIVE  POSITIONS  OF  STRAIGHT  LINES  AND 
CIRCUMFERENCES. 


PROPOSITION  I. 

1^08.  TJieorein, — Any  diameter  divides  a  circle,  and  also  its 
circumference,  into  two  equal  parts. 


Dem. — Let  AB  be  any  diameter  of  the 
circle  kmBii;  then  is  the  figure  kmB  equal  to 
A;iB. 

For  revolve  A?iB  upon  AB  as  an  axis  until  it 
falls  on  the  plane  of  kruB.  Then,  since  every 
point  in  A?iB  is  at  the  same  distance  from  the 
centre  C,  as  eveiy  point  in  ^mB,  the  figures 
vt^ill  coincide,  and  are,  consequently,  equal. 
Hence  surface  kiiB  =  surface  A??iB,  and  arc 
A/iB  =  arc  A//?B.     q.  e.  d. 


PROPOSITION  IL 

ISO,  TJieorem, — A  radius  wTiich  is  perpe7idicular  to  a  clioi'd 
Usects  the  chord  and  also  the  subtended  arc, 

Dem. — Let  AB  be  any  chord  and  OE  a  radiu? 
perpendicular  to  it  at  D;  then  AD  =  BD,  and 
AE  =  BE  * 

For,  drawing  the  radii  OA  and  OB,  revolve 
the  semicircle  CBE  upon  the  diameter  CE  until 
it  falls  on  CAE.  The  semicu-cles  will  coincide 
{158)  \  and  since  AB  is  perpendicular  to  OE, 
DB  will  fall  in  DA.  Moreover,  as  there  cannot 
be  two  equal  oblique  lines  fi'om  a  point  to  a  line 
on  the  same  side  of  a  perpendicular,  OB  and  OA 
must  coincide.  Hence  BD  coincides  with  AD, 
and  BE  with  AE.  Therefore  AD  =  BD,  and  AE 
=  BE.    Q.  e.  d. 

*  To  avoid  confusing  the  pupil  by  a  multiplicity  of  details,  the  demonstration?  in  this  sec- 
tion are  generally  limited  to  the  consideration  of  arcs  less  than  a  semi-circumference.  All  the 
propositions,  except  Prop.  V.,  are  equally  true  whatever  the  arcs,  and  the  demonstrations  can 
easily  be  applied  to  cases  in  which  the  arcs  are  greater  than  semi-circumferences.  But  this  ha4 
better  not  be  done  till  a  review  is  taken,  for  the  reason  given  above. 


Fig.  132. 


STRAIGHT   LINES   AND   CIRCUMFERENCES. 


79 


160.  Cor.  1. — A  radius  loliicli  is  perpendicular  to  a  cliord  bisects 
the  angle  suUended  hy  the  arc  of  that  chord. 

Thus  OE  bisects  AOB,  since  BOE  is  found  to  coincide  witk  AOE  in  the 
demonstration  above. 

ISI.  Cor.  2. — Conversely,  A  radius  which  iisects  an  arc  is  per- 
pe7idicular  to  the  chord  of  that  arc  at  its  middle  2^oint. 

Dem. — If  OE  bisects  arc  AB  at  E,  when  semicircle  CBE  is  revolved  on  CE 
till  it  falls  on  CAE,  EB  will  coincide  with  EA ;  and  as  D  remains  fixed  and  B 
falls  on  A,  BD  coincides  with  DA.  Hence  OE  has  two  points,  0  and  D,  each 
equidistant  from  the  extremities  of  AB,  and  is,  consequently,  perpendicular  to 
it  at  its  middle  point. 

162,  Cor.  3. — Also,  conversely,  A  radius  ivhich  bisects  a  chord  is 
perpendicular  to  the  chord  and  bisects  the  subtended  arc. 

For  it  has  two  points,  each  equidistant  from  the  extremities  of  the  chord. 

163,  Cor.  4. — The  li7ie  OD  measures  the  distance  of  the  chord  AB 
from  the  centre;  since  by  the  distance  from  a  point  to  a  line  is 
always  meant  the  shortest  distance. 


PROPOSITION   III. 

16d,  Theorem. — In  the  same  or  in  equal  circles,  equal  chords 
are  equally  distant  from  the  centre. 

Dem.— Let  0  and  0' 
he  two  equal  circles,  and 
chord  EF=  chord  CH; 
then  are  the  perpendicu- 
lars LO  and  NO', which 
measure  the  distances 
of  the  chords  from  the 
centre  {163),  equal. 

For,  since  FE  is  per- 
pendicular to  LO  and 
CHtoNO',andLF  =  NH  Fig.  133. 

(159),  the  equal  obhque  lines  FO  and  HO'  cut  off"  equal  distances  from  the  foot 
of  each  perpendicular  {141).    Therefore  LO  =  NO',    q.  e.  d. 


80 


ELEMENTARY  PLANE  GEOMETRY. 


PROPOSITION  IT. 

lOo,  Tlieoreni. — I?i  the  same  or  in  equal  circles,  equal  arcs  have 
equal  chords;  and  conversely,  equal  chords  suhtend  equal  arcs. 

Dem.— Let  0  and  0'  be 

the  centres  of  two  equal 
circles,  and  arc  kmB  —  arc 
C;iD;  tlien  chord  AB  = 
chord  CD. 

Apply  the  circle  O'  to 
the  circle  0,  with  0'  at  0, 
and  C  at  A.  Since  the  cir- 
cumferences coincide,  all 
the  points  in  each  being 
equall}'  distant  from  the 
centre,  and  since  arc  AwB  =  arc  CnD  by  hypothesis,  D  will  fall  at  B.  Hence 
AB  =  CD. 

Convei-sely,  if  chord  AB  =  chord  CD,  arc  A;wB  =  arc  Cr^D.  Draw  the  per- 
pendiculars OL  and  O'N  from  the  centres  to  the  chords.  Conceive  the  plane 
of  circle  0'  placed  upon  circle  p,  so  that  CD  shall  fall  upon  its  equal  AB,  and 
O'  be  on  the  same  side  of  AB  as  0.  Since  L  and  N  are  the  middle  points  of 
the  equal  chords,  they  will  coincide;  and  as  LO  and  NO'  are  perpendiculars  to 
the  respective  chords,  and  equal  {164),  0'  will  fall  at  0.  As  the  circles  are 
equal,  the  circumferences  will  coincide,  and  consequently  the  arc  kmB  coin- 
cides with  CwD. 


Fig.  134. 


PROPOSITION  T. 

166*  Tlieoreni, — In  the  same  or  in  equal  circles,  the  less  of 
two  arcs  has  the  shorter  chord ;  and,  conversely,  the  shorter  chord 
subtends  the  less  arc. 

Dem.— Let  0  and  0'  be  the  centres  of  two  equal  circles,  and  arc  AwB  be  less 

than  arc  C/iD;  then  is 
chord  A3  <  chord  CD. 
Drawing  the  diameters 
AL  and  CM,  place  circle 
O'  upon  circle  O,  with 
CM  upon  AL.  Take  arc 
AD'  =  arc  CiiD  and 
draw  AD',  OB,  and  OD'. 
AD'  =  CD  by  {W5), 
Now  AB  <  AN  +  NB. 
AlsoOD'<ND'  +  NO; 
Fig-  135.  or,  as  OD'  =  OB,  OB  < 

ND'  +  NO.    Subtracting  NO   from  both  members,  OB  —  NO  (or  NB)  <  ND'. 

Hence,  we  may  substitute  ND'  for  NB  in  the  inequality  AB  <  AN  +  NB  and 

have  AB  <  AN  4-  ND'  or  AD',  which  equals  CD. 


STKAIGHT  LINES   AND   CIKCUMFEEENCES.  81 

Conversely,  if  chord  AB  is  less  than  chord  CD,  arc  AwB  is  less  than  arc 
CnD.  For  if  arc  AmB  =  arc  CnD,  cliord  AB  =  chord  CD  [1(S5).  And,  if  arc 
AmB  >  arc  CwD,  chord  AB  >  chord  CD.  But  both  of  these  conclusions  are 
contrary  to  the  hypothesis.  Hence,  as  arc  AmB  can  neither  be  equal  to  not 
greater  than  arc  CtiD,  it  must  be  less. 


PROPOSITION  VI. 

167*  TJieorem, — In  the  same  or  in  equal  circles,  of  tivo  unequal 
cliords,  tlie  less  is  at  the  greater  distance  from  the  centre. 

Dem. — Let  CE   <  AB,  then  is  the  perpendicular  OD,  which  measures  the 
distance  of  CE  from  the  centre,  greater  than   OD' 
which  measures  the  distance  of  AB  from  the  centre. 

From  A  lay  off  AE'  =:  CE,  and  draw  the  perpen- 
dicular OD".  Then  OD"  =  OD,  since  equal  chords 
are  equally  distant  from  the  centre.  As  arc  AE'  < 
arc  AB,  AB  cuts  OD"  in  some  point  as  H.  NoV  OH 
>  OD'  since  the  former  is  oblique  and  the  latter  per- 
pendicular to  AB.  Also  OD"  >  OH.  Much  more 
then  is  OD"  >  OD'.      Therefore  OD  (which   equals 

OD")  >   OD'.     Q.   E.  D.  Fig.  136. 

168,   Cor.— Conversely,    Of   tiuo    chords  which   are   unequalhj 
distant  from  the  centre,  that  which  is  at  the  greater  distance  is  the 


Dem. — Thus,  if  CE  is  at  a  greater  distance  from  the  centre  than  AB,  CE  < 
AB.  For,  if  CE  were  equal  to  AB,  it  would  be  equally  distant  from  the  centre. 
And  if  CE  were  greater  than  AB,  it  would  be  at  a  less  distance  from  the  centre. 
Hence,  as  CE  cannot  be  at  an  equal  distance  from  the  centre  with  AB,  nor  at  a 
less  distance,  it  miist  be  at  a  greater. 


PROPOSITION  vn. 

169.  Theorem.^A  straight  line  can  intersect  a  circumference 
in  only  two  points. 

Dem.— The  distances  from  the  centre  to  the  intersections,  being  radii,  are 
equal.  Hence,  as  there  can  be  only  two  equal  straight  linos  drawn  from  ai 
point  to  a  straight  line,  there  can  be  only  two  intersections.    Q.  e.  d. 

0 


82 


ELEMENTARY   PLANE  GEOMETRY. 


PROPOSITION  vin. 

170,  Tlieorem, — A  straight  line  which  intersects  a  circumfer- 
e7ice  in  one  j^oint  intersects  it  also  in  a  secojid  2)oint. 

Dem. — Let  LM  intersect  the  circumference  at  A ; 
then  does  it  intersect  at  some  other  point,  as  B. 

For,  since  LM  intersects  the  circumference,  it 
passes  within  it,  and  has  points  nearer  to  0  than  A. 
The  radius  OA  is,  therefore,  an  oblique  line.  Now 
two  equal  oblique  lines  can  be  drawn  from  0  to  the 
straight  line  LM.  But  all  points  in  the  plane  at  the 
distance  OA  from  0,  are  in  the  circumference.  Hence 
there  is  a  second  point,  as  B,  common  to  LM  and  the 
circumference,     q.  e.  d. 

Fig.  ]3T. 


171*  Cor. — Any  line  which  is  oUique  to  a  radius  at  its  extremity, 
is  a  seca7it  line.  0 


PROPOSITION  IX. 

172,  Theorefii, — A  straight  line  tvhich  is  perpendicular  to  a 
radius  at  its  extremity  is  tangent  to  the  circumference, 

Dem. — The  line  touches  the  circumference  because  the  extremity  of  the 
radius  is  in  the  circumference.  Moreover,  it  does  not  intersect  the  circum- 
ference, since,  if  it  did,  it  would  have  points  nearer  the  centre  than  the  extremity 
of  the  radius;  but  these  it  cannot  have,  as  the  perpendicular  is  the  shortest 
distance  from  a  point  to  a  line.  Hence,  as  a  line  which  is  perpendicular  to  a 
radius  at  its  extremity  touches  the  circumference  but  does  not  intersect  it,  it  is 
a  tangent  {53).    Q.  e.  d. 


173.  Cor. — ConTersely,  A  tangent  to  a  circumference  is  perpen- 
dicular to  a  radius  at  the  poinl  of  contact. 

For,  as  a  tangent  to  a  circumference  does  not  pass  within,  the  point  of  contact 
is  the  nearest  point  to  the  centre,  and  hence  i*  the  foot  of  a  perpendicular  from 
the  centre. 


STRAIGHT  LINES  AND   CIRCUMFERENCES. 


83 


PROPOSITION  X. 


174,  Theorem, — Two  parallel  secants  intercept  equal  arcs. 

Dem.— Let  the  parallels  LM  and  RS  intersect  the  circumference  AECF  ;  tlien 
Hre  the  intercepted  arcs  AB  and  DC  equal. 

Draw  the  diameter  EF  perpendicular  to  one 
of  the  parallels,  as  LM,  whence  it  will  be  per- 
pendicular to  the  other  {144).  Draw  the  radii 
OB  and  OD.  Revolve  the  portion  of  the  figure 
on  the  right  of  EF,  upon  EF  until  it  falls  on  the 
plane  on  the  left  of  EF.  Then,  since  RS  and 
LM  are  perpendicular  to  EF,  IS  will  fall  in  IR, 
and  HM  in  HL.  Moreover,  as  there  cannot  be 
two  equal  oblique  lines  on  the  same  side  of  a 
perpendicular,  and  from  the  same  point  {140), 
OD  and  OB  must  coincide,  and  D  fall  at  B.  In  like  manner  C  falls  at  A,  and 
CD  coincides  with  AB.     Therefore  CD  :=  AB.    q.  e.  d. 


(V 


PROPOSITION   XI. 

17S,  Theorein* — If  a  secant  le  parallel  to  a  tangent^  tlie  arcs 
intercepted  hetioeen  the  intersections  and  the  point  of  tangency  are 
equal, 

Dem. — Let  the  secant  LM  be  parallel  to  p 

the  tangent  RS  ;  then  is  CP  -  EP. 

For,  draw  the  radius  OP  to  the  point  of 
tangency ;  it  will  be  perpendicular  to  the 
tangent  {173),  and  also  to  the  parallel 
LM  {144).  But  a  radius  which  is  perpen- 
dicular to  a  chord,  as  OP  to  CE,  bisects  the 
subtended  arc  {159),  hence  CP  —  EP.  In 
like  manner,  if  VU  is  parallel  to  LM, 
CB  =  EB.     Q.  E.  D.  ^  ^  ^ 

^  Fig.  139. 

176*  Cor. — Tioo  parallel  tangents  inchide  eq^ial  arcs  hetween  the 
points  of  tangency  ;  and  these  arcs  are  semi-circumferences» 


R              /-^ 

~"~\             s 

L  0/              D 
0 

\e  m 

64  ELEMENTARY   PLANE   GEOMETRY. 

EXERCISES. 

1.  Draw  a  circle  ana  divide  it  into  two  equal  parts.  What  proposi- 
tion is  involved  ? 

2.  Given  a  point  in  a  circumference,  to  find  where  a  semi-circum- 
ference reckoned  from  this  point  terminates.  What  proposition  is 
involved  ? 

3.  JProb, — To  bisect  a  given  arc. 
OL  B 

"^  ^'--.^  A  Solution. — Let  AB  be  an  arc  which  wo.  wish  to 

"\      /     I  bisect.*    Draw  its  cliord  AB,  and  erect  00'  bisecting 

the  chord,  by  {130).  Now,  as  00'  is  perpendicular 
to  the  chord  at  its  middle  point,  it  bisects  the  arc  by 
{162),  since  there  can  be  but  one  perpendicular  at  the 
middle  point  of  the  chord.  The  arc  AB  is,  therefore, 
Fig.  140.  bisected  at  0,  i.  e.,  AC  =  CB. 

4.  Proh, — To  bisect  a  given  ajigh. 

SuG. — The  method  of  solving  this  is  given  in  Part  I.  The  student  should 
do  it  as  there  directed,  and  then  point  out  the  principle  upon  which  the  method 
depends. 

5.  In  a  circle  whose  radius  is  11  there  are  drawn  two  chords,  one 
at  6  from  the  centre,  and  one  at  4.  Which  chord  is  the  greater  ?  By 
what  proposition  ? 

6.  In  a  certain  circle  there  are  two  chords,  each  15  inches  in 
length.  What  are  their  relative  distances  from  the  centre  ?  Quote 
the  principle. 

7.  There  is  a  circular  plat  of  ground  whose  diameter  is  20  rods. 
A  straight  path  in  passing  runs  within  ^  rods  of  the  centre.  What 
is  the  position  of  the  path  with  reference  to  the  plat  ?     What  is  the 

position  of  a  straiglit  path  whose  nearest 

hy^^  point  is  10  rods  from  the  centre  ?     One 

.'-'''    ^^  X  whose  nearest  point  is  11  rods  from  the 

/      y^  yp        centre? 

y^\  X  8.   Pass  a  line  through  a  given  point, 

A.     \      y^  and  parallel  to  a  given  line,  by  the  prin- 

y^P  ciples  contained  in  (174)  and  {165). 

Fig.  141. 


*  Thi?  solntion  and  many  others  are  given,  not  so  mnch  that  it  is  feared  that  the  student 
will  not  be  able  to  «olve  the  problem?,  as  to  afford  models  for  describing  the  process.  In 
this  case  an  arc  should  be  drawn  first,  and  all  trace  of  the  centre  obliterated.  Then  proceed  aa 
directed. 


STRAIGHT  LINES  AND   CIRCUMFERENCES. 


85 


9.  Prob, — To  draw  a  tangent  to  a  circle 
at  a  given  point  in  the  circumference. 

Solution. — Let  P  be  the  point  at  which  a  tan- 
gent is  to  be  drawn.  Draw  the  radius  OP  to  the 
given  point  of  tangency,  and  produce  it  any  con- 
venient distance  beyond  the  circle.  Erect  a  per- 
pendicular to  this  line  at  P,  as  MT ;  then  is  MT  a 
tangent  to  the  circle  {172). 


Fig. 141*. 

10.  I?r6h. — To  find  the  centre  of  a  circle  whose  circumference  is 
hnoiun,  or  of  any  arc  of  it, 

SuG. — The  pi'ocess  is  given  in  Pakt  I,     Do  the  work  as  there  directed,  and 
then  show  upon  what  proposition  in  this  section  it  is  founded. 


w    . 

O    CO 

<  o 

o  ^ 
m  ^ 
S  ^ 
%^ 

03 

o 

Oh 

> 
H 

W 


SYNOPSIS. 

Diameters.       Prop.  I.     How  divide  circles  and  circumferences. 


Chords. 


Secants. 


Tangents. 


Parallels. 


Exercises. 


r 


Prop.  II. 


Prop.  III. 
Prop.  IV. 
Prop.  V. 
Prop.  VI. 


Bisects  angle. 
Converse  of  Prop. 


r  Cor.  1. 
Radius  perp.  J  Cor.  2. 
to  chord.  I  Cor.  3.  "         *'       " 

[  Cor.  4.  Dist.  from  centre. 
Distance  of  equal  chords  from  centre. 
Equal  arcs,  and  converse. 
Unequal  arcs. 

Unequal  chords.     Dis-  ^  ^       Converse, 
tance  from  centre.     '^ 


Prop.  VII.   Intersect  in  only  two  points. 
Prop.  VIII.  If  a  lin^e  intersect  m  one  ^  ^^^  ^ine  oblique 
^rlnanoTh^r'^J    to  radius  at  extr. 

Prop.  IX.     Line  perpendicular  to  )  ^      Converse, 
radius  at  extremity.    ) 


Prop.  X. 

Prop.  XI.    Secant  par.  to  tangent. 


Parallel  secants  intercept  equal  arcs. 

Cor.  Two  parallel 
tangents. 

r  Proh.  To  bisect  an  arc. 

'   Proh.  To  bisect  an  angle. 

Prob.  To  draw  a  tangent  at  a  point  in  circumference. 

Proh.  To  find  centre  of  circumference  or  arc. 


86  ELEMENTABY  PLANE  GEOMETRY. 


SECTION   V, 

OP  THE  RELATIVE  POSITIONS  OF  CIRCUMFERENCES. 


PROPOSITION  L 

177,  Theorem, — All  the  circumferences  wTiich  may  le  passed 
tlirougli  three  ])oints  not  in  the  same  straight  line  coi^icide,  and  are 
one  and  the  same. 

Dem. — Let  A,  B,  and  C  be  three  points  not  in  the  same  straight  line ;  then 
all  the  circumferences  which  can  be  passed  through  them  will  coincide. 

For  join  the  points,  two  and  two,  by  straight  lines,  as  AB  and  BO.    Bisect 
these  lines  with  perpendiculars,  as  DF  and  EH.     Since 

..■ p."""--.,  AB  and  BC  are  not  in  the  same  straight  line,  DF  and 

/         \  Q      EH  will  meet  when  sufficiently  produced,  at  one  and 
•  /■-.       only  one  point,  as  O,  because  they  are  straight  lines 

I      -f /  j       Now,  every  point  in  FD  is  equally  distant  from  A  and 

^^  /  7^/       B,  and  eveiy  pomt  in  HE  is  equally  distant  from  B  and 

V"---:.._        /  /        c  {129).    Hence  0  is  equally  distant  from  the  three 
X^  ,.--'' B         points  A,  B,  and  C;  and,  if  a  circumference  be  drawn 

'"  with  0  as  a  centre,  and  a  radius  AG,  it  will  pass  through 

the  three  points.  Moreover,  every  circumference  pass- 
ing through  these  points  must  have  O  for  its  centre,  since  the  centre  must  be  in 
FD  (otherwise  it  would  be  unequally  distant  from  A  and  B),  and  also  in  HE 
{129).  But  these  lines  intereect  only  in  O.  Also,  every  circumference  with  0 
as  its  centre,  and  passing  through  A,  must  have  AG  for  its  radius.  Hence,  as 
all  circles  having  the  same  centre  and  the  same  radius  coincide,  all  those  passing 
through  three  poiats.  A,  B,  and  C,  coincide.    Q.  e.  d. 

178,  Cor.  1. — Tlirough  any  three  points  not  in  the  same  straight 
line  a  circumference  can  le  passed. 

179,  Cor.  2. — T7iree  poi7its  not  in  the  same  straight  line  determine 
n  circumference  as  to  position  and  extent ;  i.  e.,  in  all  respects. 

ISO,  Cor.  3. — Tiuo  circumferences  can  intersect  in  07ily  tioo 
points. 

For,  if  they  have  three  points  common,  they  coincide,  and  form  one  and  the 
same  circumference. 


RELATIVE  POSITIONS  OF  CIRCUMFERENCES. 


87 


PROPOSITION  II. 

181,  TlieQvem. — Two  circumferen- 

ces  ivMch  intersect  in  one  point,  intersect 
also  in  a  second  point. 

Dem. — Let  M  intersect  N  at  P.  As  M  passes 
from  without  to  witliio  the  circle  N,  it  has  points 
both  without  and  within.  Now,  for  M  to  re- 
turn into  itself  from  any  point  within  N,  as  Y, 
to  any  point  without,  as  X,  it  must  intersect  N ; 
but  it  cannot  intersect  in  P,  for  a  circumference 
dDcs  not  intersect  itself  Hence,  it  intersects  in 
I  second  point,  as  P'.    q.  e.  d. 


Fig.  143. 


PROPOSITION  in. 

182.  TJieorem, — If  a  straight  line  he  drawn  through  the  cen- 
tires  of  two  circles,  of  the  intersections  of  either  circumference  luith 
tlat  line,  the  one  on  the  side  toward  the  centre  of  the  other  circle  is 
tie  nearest  jmnt  in  this  circumference  to  that  centre,  and  the  one  on 
tie  opposite  side  is  the  farthest  point  from  that  centre. 

Dem. — Let  M  and  N,  or  M'  and  N',  be  two  circumferences  whose  centres  are 
C  and  O'.    Draw  an  indefinite  line  through  these  centres.    Let  A  and  H  be  the 
intersections  of  M  or  M'  with  this  line,  of  which  A  is  on  the  side  of  M  or  M' 
toward    the  centre  O',   and 
H  is  on  the  opposite  side. 
Then  is  A  the  nearest  point 
in  M  or  M'  to  0',  and  H  the 
farthest  point  from  0'. 

Mrst,  To  show  that  A  is 
nearer  0'  than  any  other 
point  in  the  circumference. 
A  will  lie  between  0  and  0', 
in  0',  or  beyond  0'.    When 

A  lies  between  O  and  0',  as  in  M,  let  P  be  any  other  point  in  M,  and  draw  OP 
and  O'P.  Now  00'  being  a  straight  line,  is  less  than  OPO',  a  broken  line. 
Subtracting  OA  from  the  former,  and  its  equal  OP  from  the  latter,  we  have  AO' 
<  PO'.  When  A  falls  at  0'  the  truth  is  self-evident.  When  A  lies  beyond  O', 
as  in  M',  let  P  be  any  other  point  in  M',  and  draw  OP  and  O'P.  Now  O'P  + 
00'  >  OP  (=  OA).  Subtracting  00'  from  both,  we  have  O'P  >  OA  —  00' 
(=  O'A).     Hence,  in  any  case,  A  is  the  nearest  point  in  M  or  M'  to  0'. 

Second,  To  show  that  H  is  the  farthest  point  in  M  or  M'  from  0'.    In  either 


Fig.  144. 


88 


ELEMENTARY  PLANE  GEOMETRY. 


figure,  let  P'  be  any  other  point  in  the  circumference  than  H,  and  draw  OP 
and  O'P'.  >'ow,  PO  +  00'  >  P'O'.  But  P'O  =  HO.  Hence  HO  +  00'  (= 
HO')  >  P'O' 


PROPOSITION  IT. 

183.  Theorem. —  When  tlie  distance  between  the  centres  of  two 
circles  is  greater  than  the  sum  of  their  radii,  the  circumferences  are 
wholly  exterior  the  one  to  the  other. 

Dem. — Let  M  and  N  be  the  circumferences  of  two  circles  whose  centres  ar« 

O  and  O'.  Let  00'  be  greater  than  the 
sum  of  the  radii.  Then  are  M  and  N 
wholly  exterior  the  one  to  the  other.  / 
For  A,  the  intersection  of  M  witi 
00',  is  between  O  and  0',  since  OA  <" 
00',  Now,  by  hj-pothesis,  00'  >  OA 
+  BO'.  Subtracting  OA  from  both,  \ge 
have  AO*  >  BO'.  Hence,  as  the  neareJt 
point  in  M  is  farther  from  O'  than  the 
circumference  of  the  latter  circle,  M  lies  wholly  exterior  to  N.     q.  e.  d. 


Fig.  145. 


184,  Cor. — Conversely,  W7ie7i  two  circumferences  are  exterior  th 
one  to  the  other,  the  distance  between  their  centres  is  greater  than  tliB 
sum  of  their  radii. 

Dem. — For,  join  the  centres  00'  with  a  straight  line.  Now  tlie  point  A 
where  this  line  cuts  the  circumference  M  is  the  nearest  point  in  this  circumfer- 
ence to  the  centre  0'.  But,  by  hypotliesis,  this  (and  every  other  point  in  cir- 
cumference O)  is  without  circle  O'.  Hence,  AO'  >  BO'.  To  each  add  OA, 
and  OA  +  AO'  (or  00')  >  OA  +  BO'. 


PROPOSITION  T. 

185.  Tlieoreni, — When  the  distance  hetiueen  the  centres  of  two 
circles  is  equal  to  the  sum  of  their  radii,  the  circumferences  are  tan- 
gent to  each  other  externally. 

Dem. — Let  M  and  N  be  two  circumferences,  and  00',  the  distance  between 
their  centres,  be  equal  to  00  +  O'C,  the  sum  of  their  radii;  then  are  the  cir- 
cumferences tangent  to  each  otlier  externally. 


RELATIVE   POSITIONS   OF   CIRCUMFERENCES. 


89 


The  point  A,  where  M  cuts  the  line  join- 
ing the  centres,  is  between  0  and  0',  since 
OA  <  00'  by  hypothesis.  Moreover,  A 
is  the  nearest  point  in  M  to  the  centre  0'. 
Again,  as  00'  =  00  +  O'C,  subtracting  OA 
from  the  first  member,  and  its  equal  00  from 
the  other,  we  have  O'A  =  O'C;  that  is,  A  is  in 
the  circumference  N.  Hence,  as  A  lies  in  N, 
and  all  other  points  in  M  are  more  distant 
from  0'  than  the  length  of  the  radius  O'C,  M 
is  entirely  without  N,  except  the  point  A,  and  the  circles  are 
other  externally,     q.  e.  d. 


to  each 


ISO,  Cor.  1. — Conversely,  Wlien  two  circumferences  are  tangent 
to  each  other  externally,  the  distance  betiveen  their  centres  is  equal  to 
the  sum  of  their  radii. 

Dem. — M  being  tangent  to  N  externally,  the  point  in  M  nearest  the  centre  O' 
must  be  in  N,  while  all  other  points  in  M  are  exterior  to  N.  Now,  the  point  in 
M  nearest  to  0'  is  A  on  the  line  joining  their  centres  {182).  A  is  therefore  the 
point  of  tangency,  and  00'  =  OA  +  O'A. 

187 »  Cor.  2. —  When  two  circumferences  are  ta7igent  to  each  other 
externally,  the  point  of  tangency  is  in  the  line  joining  their  centres. 


PROPOSITION  VI. 


L- 


188,  Uieorem, —  WJie^i  the  distance  hetiueen  the  centres  of  two 
circles  is  less  than  the  sum  and  greater  than  the  difference  of  their 
radii,  the  tivo  circumferences  intersect. 

Dem. — Let  M  and  N  be  the  circumferences  of  two  circles  whose  centres  are 
O  and  0'.    Let  the  radius  of 

M  be  equal  to  or  greater  than  ^ — --\  N 

the  radius  of  N,  Now,  if  00' 
<0A  +  O'B,  and  >  OA— O'B, 
M  and  N  intersect. 

For,  when  00'  >  OA,  00' 
<  OA  +  O'B  gives  00'  —  OA 
(=    AO')  <   O'B  ;    and  when  Fig.  147. 

00'  <  OA,  00'  >  OA  -  O'B  gives  O'B  >  OA  -  00'  (=  AO')-  Hence  the 
nearest  point  in  M  to  0'  lies  within  N.  Again,  to  the  first  member 
of  00'  >  OA  —  O'B  add  HO,  and  to  the  second  its  equal  OA,  and  we 
have  00'  -I-  HO  (=:  HO')  >  20A  -  O'B.  Now,  since  O'B  ^  OA,*  by  hypothesis, 
the  difference  20A  -  O'B  ^  O'B.  Hence,  HO'  >  O'B,  and  H  lies  without  N.    As, 


♦  Read  "  O'B  is  equal  to  or  less  tliau  0A-' 


90  ELEMENTARY  PLANE  GEOMETRY. 

therefore,  M  has  oue  pomt  at  least  within  N  and  one  without,  M  and  N  inter- 
sect.    Q.  E.  D. 

189,  CoE. — Conversely,  WJien  tiuo  circumferences  intersect,  the 
distance  heticeen  their  centres  is  less  than  the  sum  and  greater  than 
the  difference  of  their  radii. 

Dem. — Let  the  radius  of  N  be  equal  to  or  less  than  the  radius  of  M.  As  the 
circumferences  intersect  the  farthest  point  H'  of  N  from  0  must  be  farther  from 
O  than  the  length  of  the  radius  of  M,  i.  e.,  must  lie  without  that  circle.  So  we 
have  by  hypothesis  H'O  >  OA.  Subtracting  H'O'  from  the  first  menil)er  and  its 
equal  BO'  from  the  second,  we  have  H'O  —  O'H'  (=  00')  >  OA  -  BO';  that  is, 
the  distance  between  the  centres  is  greater  than  the  difi'erence  of  the  radii. 
Again,  as  the  nearest  point  in  M  to  O'  must  lie  within  N,  we  have  AO'  <  BO', 
and  adding  OA  to  both  members,  OA  +  AO'  (=  00')  <  OA  +  BO';  that  is,  the 
distance  between  the  centres  is  less  than  the  sura  of  the  radii. 


PROPOSITION  TIL 

190,  Theorem. —  When  the  distance  hetween  the  centres  of  two 
unequal  circles  is  equal  to  the  difference  of  their  radii,  the  less  cir- 
cumference is  tangent  to  the  other  internally. 

Dem.— Let  M  and  N  be  the  circumferences  of  two  circles  whose  centres  0  and 
O'  are  so  situated  that  00'  =  00  -  O'C ;  then  are  the 
circles  tangent  to  each  other  internally. 

For,  let  N  be  the  circumference  of  the  less  circle,  so 
that  00  >  O'C  Let  HH'  be  a  diameter  of  M.  Bj 
hypothesis  00'  =  00  —  O'C.  Now,  subtracting  each 
member  of  this  eq  tj,lity  from  OH',  we  have  OH'  —  00' 
(=  O'H')  =  O'C.  Whence  it  appears  that  H',  the  point 
in  N  at  the  greatest  distance  from  O,  is  in  M ;  and,  con- 
FiG.  148.  sequently,  that  every  other  point  in   N   is  within   M. 

Hence,  N  is  tangent  to  M  internally.     Q.  e.  d. 

191.  Cor.  1. — Conversely,  Wlien  a  less  circumference  is  tangeiit  to 
a  greater  internally,  the  distance  hetween  their  centres  equals  the 
difference  of  their  radii. 

Dem. — The  less  circumference  N  being  tangent  to  the  greater  M,  internally, 
the  point  in  N  at  tlie  greatest  distance  from  the  centre  0  of  M,  must  be  in  M, 
while  all  other  points  of  N  lie  within  M.  Now  H'  in  the  line  passing  through 
the  centres  is  the  point  of  N  at  the  greatest  distance  from  O.  Hence  we  ob- 
serve  that  00'  -  OH'  -  O'H',  i.  e.,  the  difi'erence  between  the  radii. 


RELATIVE   POSITIONS  OF  CIRCUMFERENCES. 


91 


192.  Cor.  2. —  When  one  circumference  is  ta?ige?it  to  another  in- 
iernally^'the  i^oint  of  iangency  is  in  the  line  passi?ig  through  their 
centres. 

193,  ScH. — If  the  radii  are  equal  the  two  circumferences  coincide. 


PROPOSITIOX  Tin, 

194.  Tlieore'in,  —  When  the  distance  ietiueen  the  centres  of  two 
unequal  circles  is  less  thari  the  difference  of  their  radii,  the  less  cir- 
cumference is  wholly  within  the  greater. 

Dem. — Let  N  be  a  less  circumference  than  M,  and  00', 
the  distance  between  their  centres,  be  less  than  OA  — O'H', 
the  difference  of  their  radii ;  then  is  N  wholly  within  M. 

For,  to  each  member  of  00'  <  OA  —  O'H'  add  O'H',  and 
wehaveOO'  +  0'H'<OA.  But  00'  +  0'H'=  OH'.  Hence 
0H'<  OA,  and  H',  the  farthest  point  in  N  from  0,  is  with- 
in M,  and  consequently  N  lies  wholly  within  M,    o.  e.  d. 

Fig.  149. 

195.  Cor. — Conversely,  Wlien  a  less  circumference  is  wholly 
within  a  greater,  the  distance  hetiveen  their  centres  is  less  than  the 
difference  of  their  radii. 

Dem.— If  N  lies  wholly  within  M,  the  farthest  point  in  N  from  O,  the  centre 
of  M,  must  be  nearer  0  than  is  any  point  iu  M,  i.  e.,  0H'<  OA.  Now,  subtract 
O'H'  from  each  member,   and  we   have   OH'  —  O'H'  {-—  00')  <  OA  —  O'H'. 

Q.   E.    D. 

196.  ScH.— If  the  centres  coincide  so  that  00'  =  0,  the  circumferences  are 
said  to  be  concentric.  If,  at  the  same  time,  their  radii  are  equal,  they  are  coin- 
cident. 


PROPOSITION  IX. 

197.  TJieorem. —  Wlien  two  circumferences  intersect,  the  line 
which  passes  through  their  centres  is  per- 
pendicular to  their  common  chord  at  its 
middle  point. 

Dem.— Let  the  circumferences  M  and  N 
intersect -in  the  points  P  and  P'  {181);  let 
^P'  be  the  common  chord,  and  LR  the  line 
l^assinc:  tlirough  the  centres  0  and  0';  then  is 
LR  perpendicular  to  PP'. 


92  ELEMENTARY  PLANE  GEOMETRY. 

For  0'  is  equally  distant  from  the  extremities  P  and  P',  and  0  is  also  equally- 
distant  from  P  and  P'.  Hence,  as  LR  has  two  points  equally  distant  from  the 
extremities  of  PP',  it  is  perpendicular  to  PP'  at  its  middle  point,     q.  e.  d. 


PROPOSITION  X. 

19S»  Tlieorem. —  When  one  circumference  is  tangent  to  another, 
either  externally  or  internally,  they  have  a  common  rectilinear"^  tan- 
gent at  their  common  point. 

Dem. — Since  the  radii  of  the  two  circles  drawn  to  the  common  point  form 
one  and  the  same  straight  line  {187 f  102),  a  line  perpendicular  to  one  at  its 
extremity  is  perpendicular  to  the  other  also.  And  a  line  which  is  perpen- 
dicular to  a  radius  at  its  extremity  is  tangent  to  the  circle  {172).    Q.  e.  d. 

199.  Cor. — All  circumferences  having  their  centres  in  the  same 
line,  and  having  hut  one  common  point,  are  tangent  to  each  other,  and 
have  a  common  rectilinear  tarigent  at  the  common  point. 


EXERCISES. 

1.  Prob. — To  pass  a  circumference  through  three  given  points 
not  in  the  same  straight  line. 

SUG. — The  process  should  be  gone  through  with  as  learned  from  Part  I., 
and  then  tlie  reasons  for  the  process  given  as  furnished  by  this  section. 

2.  To  pass  a  circumference  through  two  given  points,  whose 
center  shall  be  in  a  given  line. 

3.  JProb* — To  circumscribe  a  circumference  about  a  given  triangle, 
and  give  the  reasons  for  the  process. 

4.  The  centres  of  t^vo  -circles  whose  radii  are  10  and  7,  are  at  4 
from  each  other.  What  is  the  relative  position  of  the  circumfer- 
ences? What  if  the  distance  between  the  centres  is  17?  What  if 
20  ?     What  if  2  ?     What  if  0  ?     What  if  3  ? 

*  Straight  line. 


RELATIYE   POSITIONS    OF   CIIICUMFERENCES. 


93 


5.  Given  two  circles  0  aud 
0',  to  draw  two  others,  one 
of  whicli  shall  be  tangent  to 
these  externally,  and  to  the 
other  of  which  the  two  given 
circles  shall  be  tangent  in- 
ternally. Give  all  the  princi- 
ples involved  in  the  construc- 
tion.    Give  other  methods. 

6.  Given  two  circles  whose 
radii  are  6  and  10,  and  the  *^'^-  ^^^• 

distance  between  their  centres  20.  To  draw  a  third  circle  whose 
radins  shall  be  8,  and  which  shall  be  tangent  to  the  two  given  cir- 
cles ?  Gail  a  third  circle  whose  radius  is  2  be  drawn  tangent  to 
the  two  given  circles  ?  How  will  it  be  situated  ?  Can  one  be  drawn 
tangent  to  the  given  circles,  whose  radius  shall  be  1  ?     Why? 


o 

w 

& 
O 

P3 

M 

o 

o 
or 

o 


O 
Ph 

> 
i-i 

E-t 
<J 


SYNOPSIS. 

Prop.     I.  Through  three  points. 


(  Cor.  1. 

>ints.  •<  Cor.  2. 

(  Cor.  3. 


A  circf.  can  be  passed. 
A  circf.  determined  by. 
Intersections  of  two  circf 's. 


Prop.   II.  Two  circumferences  which  intersect  in  one  point. 

Prop.  III.  Points  in  one  circumference  nearest  to  and  farthest  from  tlie 
centre  of  another. 


> 


'  Prop.      IV.  Greater  than  sum  of  radii.  ■{  Cor.  Converse. 

Prop.        V.  Equal  to  sum  of  radii.  \  ^^''  l  ^^P^.^^f  ; 

^  {  Cor.  2.  Pomt  of  tangency. 

Prop.      VI.  Less  than  sum  and  greater  than  (  n     r^ 

difference  of  radii.  J  Cbr.  Converse. 


Prop.    VII.  Equal  to  diff. 


I  Cor. 

of  radii.  <  Cor. 

(Sch. 


Cor.  1.  Converse. 

2.  Point  of  tangency. 
Radii  equal. 


Prop.  VIII.  Less  than  diff.  of  radii. 


Cor.  Converse. 


Sch.  Concentric,  Coincident 
Prop.  IX.  Perpendicular  to  common  chord. 
Prop.    X.  Common  tangent  to  two  circles  tangent  to  each  \  q^^  rj,^  ^^i 

Exercises     -i  -^'^^^'  '^^  ^^^^  circumference  through  three  points. 

I  Prob.  To  circumscribe  a  triangle  with  a  circumference. 


94. 


ELEMENTABY  PLANE   GEOMETRY. 


SECTION  VL 

OF  THE  MEASUREMENT  OF  ANGLES. 


200»  Angles  are  said  to  be  measured  by  arcs,  according  to  the 
principles  developed  in  the  three  following  propositions. 


PROPOSITION  I. 


201.,  Theorem, — In  the  same  or  in  equal  circles,  equal  arcs 
subtend  equal  angles  at  the  centre. 


-In  the  equal  circles 


let  arc  AB  =  arc  DC;  then  will  the 
angles  O  and  0',  called  angles  at 
the  centre,  be  equal.  For,  placing 
N  upon  M  so  that  C  shall  fall 
on  0,  and  CD  on  OA,  since 
the  circles  are  equal,  D  will  fall 
on  A ;  and  since,  by  hypothesis, 
arc  DC  =  arc  AB,  C  will  fall  on 
B.  Hence,  O'C  will  coincide 
with  OB,  and  angle  O'  =  angle 
O,  because  they  coincide  when 
applied,    q.  e.  d. 


202,  Cor.  1. — Conversely,  In  the  same  or  in  equal  circles, eqtial 
angles  at  the  centre  intercept  equal  arcs. 

Dem. — If,  by  hypothesis,  angle  O'  =  angle  O,  in  the  equal  circles  M  and  N, 
arc  DC  =  arc  AB.  For,  placing  circle  N  upon  M,  so  that  O'  shall  foil  on  O, 
and  O'D  on  its  equal  OA,  D  falls  on  A,  and,  since  angle  O'  =  angle  0.  O'C  takes 
the  direction  OB,  and,  being  equal  to  it,  C  falls  on  B.  Hence,  DC  and  AB  co- 
incide and  are  equal. 

203,  Cor.  2. — A  right  angle  at  the  centre  intercepts  a  quarter  of 
a  circumference,  and  is  said  to  he  r)ieasured  hy  it.  Hence,  a  semi- 
circumference  is  the  measure  of  tioo  right  angles,  and  a  whole  circum,- 
ference  of  four. 


MEASUREMENT  OF  ANGLES. 


95 


r 


PROPOSITION  n. 


Fig.  153. 


204:.  Theorein, — In  the  same  or  in  equal  circles,  arcs  wJiich  are 
in  the  ratio  of  tiuo  luhole  7iumhers  subtend  angles  at  the  centre  ivhich 
have  the  same  ratio,  whe?ice  the  angles  are  to  each  other  as  the  arcs 
which  subtend  them. 

Dem. — In  tlie  equal  circles  M  and  N,  let  the  arcs  EF  and  IH,  which  subtend 
the  angles  0  and  O'  at  the  centre,  be  in  the  ratio  of  5  to  8 ;  then  are  the  angles 
O  and  0'  in  the  ratio  of  5  to  8, 
and  we  have 

angle  0  :  angle  O' ::  arc  EF :  arc  I H . 

For,  divide  EF  into  5  equal 
parts,  as  E«,  ab,  etc.,  then  IH  can 
be  divided  into  8  such  parts,  le, 
ef,  etc.  Draw  the  radii  Oa,  Ob, 
Oc,  etc.,  and  O'e,  O'/,  O'g,  etc. ; 
and,  since  these  partial  arcs 
are  equal,  the  partial  angles 
which  they  subtend  are  equal, 
by  the  preceding  proposition.  Now,  O  is  composed  of  5  of  these  angles,  and 
0'  of  8;  whence 

angle  0   :  angle  O'  : :  5  :  8. 
But,  a?-c    EF  :  arc    IH   ::  5  :  8. 

Hence,  the  two  ratios  being  equal,  we  have 

angle  0'   :  angle  0  :  :  arc  IH   :  arc  EF. 

As  the  same  method  could  be  pursued  in  case  the  arcs  were  to  each  other  as 
any  other  two  whole  numbers,  the  argument  is  general. 

206,  Cor. — Conversely,  In  the  same  or  in  equal  circles,  angles  at 
the  centre  ivhich  are  in  the  ratio  of  tivo  tvhole  numbers  are  to  each 
other  as  their  intercepted  arcs. 

Dem.— Thus,  let  angle  O'^be  to  angle  O  in  the  ratio  of  8  to  5.  Conceive  O' 
divided  into  8  equal  partial  angles,  then  will  0  be  divisible  into  5  such  partial 
angles.  Now,  the  partial  angles  being  equal,  their  intercepted  arcs  are  equal, 
by  the  preceding  proposition,  Cor.  1.     Whence, 

arc  IH  :  arc  EF 
But,  angle  O'  :  angle  0 
Hence,  arc    IH  :  arc  EF 

And  the  same  method  could  be  pursued  with  angles  having  the  ratio  of  any 
other  whole  numbers. 


:  8  :  5. 

:  8  :  5,  by  hypothesis. 

:  angle  0'  *:  angle  0. 


96 


ELEMENTARY  PLANE  GEOMETRY. 


PROPOsmox  m. 

206.  TJieoreni, — In  the  same  circle  or  in  equal  circles,  tvw  hi- 

conimensurahle  arcs  are 
^  to  each  other  as  the  angles 

which  they  suUend  at  the 
centre. 


Fig.  1S4. 


Dem. — In  the  equal  cir- 
cles M  and  N,  let  EF  and  IH 
be  incommeusurable  arcs. 
Now  there  is  some  arc  to 
which  EF  beai-s  the  same 
ratio  as  angle  0  to  angle  O'. 
If  that  arc  is  not  IH  let  it  be  IL,  an  arc  less  than  IH,  so  that 
angle  0  :  angle  (y  :  :  arc  EF  :  arc  IL  * 

Conceive  EF  divided  into  equal  parts,  each  of  which  is  less  than  LH,f  the  as^ 
sumed  difference  between  IH  and  IL.  Then  conceive  one  of  these  equal  parts 
to  be  applied  to  I H  as  a  measure,  beginning  at  I.  Since  the  measure  is  less 
than  LH,  at  least  one  point  of  division  must  fall  between  L  and  H.  Suppose  K 
to  be  such  a  point.  Draw  O'K.  Now,  the  arcs  EF  and  IK  are  commensurable, 
and  by  the  last  proposition 

angle  0  :  angle  lO'K  :  :  arc  EF  :  arc  IK.     But^we  assumed  that     ■ 
angle  O  :  angle  lO'H  :  :  arc  EF  :  arc  IL. 
In  these  proportions  the  antecedents  being  alike,  the  consequents  should  be  pro- 
portional, so  that 

angle  lO'K  should  be  to  angle  lO'H  :  :  arc  IK  :  arc  IL. 

But  this  proportion  is  false,  since 

angle  lO'K  <  angle  lO'H,  whereas  arc  IK  >  arc  IL.  ' 

In  a  manner  altogether  similar  (the  student  should  supply  it)  we  can  show 
that 

angle  0  is  not  to  angle  O'  :  :  arc  EF  :  any  arc  greater  than  IH. 
Hence,  as  the  fourth  term  of  the  proportion  cannot  be  less  or  greater  than  IH,  it 
must  be  IH  itself;  and 

angle  0  :  angle  O'  :  :  arc  EF  :  arc  IH,     q.  e.  d. 

207,  Cor. — Conversely,  In  the  same  or  in  equal  circles,  two  incom- 
mensurable  angles  at  the  centre  are  to  each  other  as  the  arcs  which 
they  intercept. 


*  This  ig  a  false  hypothesis,  and  the  object  of  the  argument  following  is  to  show  its 
falsity, 

+  This  can  be  done  by  snpposing  EF  bisected,  then  the  halves  bisected,  then  the  fonrths 
bisected,  and  this  process  of  bisection  continued  until  the  parts  are  each  less  than  LH. 


MEASUREMENT  OF  ANGLES.  97 

Dem. — In  the  equal  circles  M  and  N,  O  and  O'  being  incommensurable 
angles  at  the  centre,  are  to  each  other  as  the  arcs  EF  and  IH,  If  not,  let  us  sup- 
pose 

arc  EF  :  arc  IH  :  :  angle  0  :  angle  lO'L,  an  angle  less  than  0'. 

Divide  0  into  equal  partial  angles,  each  less  than  LO'H,  the  assumed  differ- 
ence between  lO'H  and  lO'L.  Also  conceive  this  angle  to  be  applied  as  a 
measure  to  lO'H,  beginning  at  O'l.  At  least  one  line  of  division  will  fall  be- 
tween O'L  and  O'H.  Let  O'K  be  such  a  line.  Now,  as  0  and  lO'K  are  com- 
mensurable, we  have  by  {205), 

arc  EF  :  arc  IK  :  :  a7igle  0  :  angle  I  O'K. 

But  by  supposition 

arc  EF  :  arc  IH  :  :  angle  0  :  angle  I  O'L. 

Therefore,  since  the  antecedents  are  the  same, 

arc  IK  should  be  to  arc  IH  :  :  angle  lO'K  :  angle  lO'L. 

But  this  is  false,  since 

arc  IK  <  arc  IH,  whereas  angle  lO'K  >  angle  lO'L. 

Whence  we  learn  that  the  fourth  term  of  the  proportion  cannot  be  less  than 
angle  lO'H.  In  a  similar  manner  it  can  be  shown  (let  the  student  do  it)  that  it 
cannot  be  greater.    Hence 4t  must  be  lO'H  itself;  and 

arc  EF  :  arc  IH  :  :  angle  O  :  angle  lO'H. 

208.  ScH. — Out  of  the  truths  developed  in  the  three  preceding  proposi- 
tions grows  the  method  of  representing  angles  by  degrees,  minutes,  and  seconds, 
as  given  in  Trigonometry  (Part  IV.,  3-6).  It  will  be  observed,  that  in  all 
cases,  if  arcs  be  struck  20it7i  the  same  radius,  from  the  vertices  of  angles  as 
centres,  the  angles  bear  the  same  ratio  to  each  other  as  the  arcs  intercepted  by 
their  sides.  Hence  the  arc  is  said  to  measure  the  angle.  Though  this  language 
is  convenient,  it  is  riot  quite  natural;  for  we  naturally  measure  a  quantity  by 
another  of  like  kind.  Thus,  distance  (length)  we  measure  by  distance,  as  when 
we  say  a  line  is  10  inches  long.  The  line  is  length  ;  and  its  measure,  an  inch, 
is  length  also.  So,  likewise,  we  say  the  area  of  a  field  is  4  acres  :  the  quantity 
measured  is  a  surface  ;  and  the  measure,  an  acre,  is  a  surface  also.  Yet,  not- 
withstanding the  artificiality  of  the  method  of  measuring  angles  by  arcs,, 
instead  of  directly  by  angles,  it  is  not  only  convenient  but  universally  used ,-  and 
the  student  must  know  just  what  is  meant  by  it.  For  example,  a  circumference 
is  conceived  as  divided  into  360  equal  arcs,  caU'ed  degrees.  Hence,  as  a  right 
angle  at  the  centre  is  subtended  by  one-fourth  of  the  circumference,  it  is  called 
an  angle  of  90  degrees.  180  degrees  is  the  measure  of  two  right  angles,  45  de- 
grees, of  half  a  right  angle,  etc.    Thus  we  get  a  i)erfectly  definite  idea  of  the; 

7 


98 


ELEMENTARY  PLANE  GEOMETRY. 


magnitude  of  an  angle  from  the  statement  of  the  number  of  degrees  which 
measure  it ;  and,  for  brevity,  the  angle  is  spoken  of  as  an  angle  of  the  same 
number  of  degrees  as  the  intercepted  arc. 


209^  An  Inscribed  Angle  is  an  angle  whose  vertex  is  in  a 
circumference,  and  whose  sides  are  chords,  or  a  chord  and  diameter, 
of  that  circumference. 

PROPOSITION  IV. 

210.  TJieorem. — An  inscribed  angle  is  measured  by  half  the 
arc  intercepted  betwee^i  its  sides. 

Dem. — Mr  St,  when  one  side  is  a  diameter.  Let  APB 
be  an  inscribed  angle,  and  PB  a  diameter;  then  is  APB 
measured  by  one-half  of  arc  AB.  For,  through  the 
centre  O,  draw  the  diameter  DC  parallel  to  the  chord 
PA;  then  COB  =  POD  {134),  whence  arc  CB  =  arc 
PD  {202),  also  COB  =  APB  {152);  and  arc  PD  =  arc 
AC  {174),  whence  PD  =  CB  =  iAB.  Now  COB  is 
measured  by  CB  {208)  \    hence  APB  is  measured  by 

Fig.  155.  CB-^AB.      Q.  E.  D. 


Second,  when  both  sides  are  chords  and  the  centre  of  ih£ 
circle  lies  between  them.  Let  APB  be  such  an  angle. 
Draw  the  diameter  PC.  Now,  by  the  preceding  part, 
APC  is  measured  by-^AC,  and  CPB  by  iCB.  Hence 
APC  +  CPB,  or  APB,  is  measured  by  ^AC  +  iCB,  or 
iAB.    Q.  E.  D.    ^  


Third,  when  both  sides  are  cliords  and  tlie  centre  lie.i 
without  the  angle.  Let  APB  be  such  an  angle.  Draw  the 
diameter  PC.  Now  APC  is  measured  by  iAC,  and  BPC 
by  I BC.  Hence  APC  —  BPC,  or  APB,  is  measured  by 
iAC-^BC,  or  ^AB.    q.  e.  d. 


Fig.  157. 


MEASUREMENT  OF  ANGLES. 


99 


211.  Cor. — In  the  same  or  equal  circles  all 
angles  inscribed  in  the  same  segment ,  intercept 
equal  arcs,  and  are  consequently  equal.  If  the 
segment  is  less  than  a  semicircle,  the  angles  are 
obtuse ;  if  a  semicircle,  right ;  if  greater  than  a 
semicircle,  acute, 

III.— In  each  separate  figure  the  angles  P  are  equal, 
for  they  are  each  measured  by  half  the  same  arc.  ...  In 
0,  each  angle  P  is  acute,  being  measured  by  \m,  which 
is  less  than  a  quarter  of  a  circumference.  ...  In  0',  each 
angle  P  is  a  right  angle,  being  measured  by  \w!  ^  which 
is  a  quadrant  (quarter  of  a  circumference).  ...  In  0", 
each  angle  P  is  obtuse,  being  measured  by  ^m",  which  is 
greater  than  a  quadrant. 

SCH. — The  converse. of  this  proposition  is  usually  taken 
for  granted  ;  i.  e.,  that  if  the  several  angles  P,  P,  etc.,  are 
equal  and  subtended  by  the  same  chord,  their  vertices  lie 
in  the  circumference.  This  is  readily  proved  rigorously 
after  the  next  two  propositions.  Thus,  if  vertex  P  were 
without,  the  angle  would  be  measured  by  ^AB  —  |^  the 
other  intercepted  arc ;  and  if  within,  by  ^AB  + 1  the  other 
intercepted  arc. 

PROPOSITION  V. 


212.  Theorem,— Any  angle  formed  by  two  chords  intersecting 
in  a  circle  is  measured  by  one-half  the  sum  of  the  arcs  intercepted 
between  its  sides  and  the  sides  of  its  vertical,  or  opposite,  angle. 

Dem.— Let  the  chords  AB  and  CD  intersect  at 
P;  then  is  APD,  or  its  equal  CPB,  measured  by 
•i(AD  +CB);  and  APC,  or  its  equal  BPD,  is  measured 
by  ^(AC  +  BD). 

For,  through  C  draw  CE  parallel  to  BA ;  whence 
ECD  =  APD  {152\  and  CB  =  EA  {174).  But  ECD  is 
measured  by  i  ED  {210\  which  equals  ^  (AD  +  EA)  = 
i(AD  +  CB). 

That  APC,  or  its  equal  BPD,  is  measured  by 
i(AC  +  BD),  appears  from  the  fact  that  the  sum  of  the 
four  angles  about  P  being  equal  to  four  right  angles,  is  measured  by  a  whole 
circumference  {208).  But  APD  +  CPB  is  measured  by  AD  +  CB;  whence 
APC  +  BPD,  or  2  APC,  is  measured  by  the  whole  circumference  minus  (AD  +  CB) ; 
that  is,  by  AC -fBD.     Then  is  APC  measured  by  ^  (AC +  BD). 


Fio.  159. 


100 


ELEMENTARY  PLANE   GEOMETRY. 


213,  Sen. — The  case  of  the  angle  included  between  two  chords  passes 
into  that  of  the  inscribed  angle  in  the  preceding  proposition,  by  conceiving  AB 
to  move  parallel  to  its  present  position  until  P  arrives  at  C  and  BA  coincides 
with  CE.  The  angle  APD  is  all  the  time  measured  by  half  the  sum  of  the  in- 
tercepted arcs;  but,  when  P  has  reached  C,  CB  becomes  0,  and  APD  becomes 
an  inscribed  angle  measured  by  half  its  intercepted  arc. 

In  a  similar  manner  we  may  pass  to  the  case  of  an  angle  at  the  centre, 
by  supposing  P  to  move  toward  the  centre.  All  the  time  APD  is  measured  by 
i(AD  +  CB);  but,  when  P  reaches  the  centre,  AD  =  CB,  and  i  (AD  +  CB)  = 
\  (2AD)  =  AD  ;  i.  <?.,  an  angle  at  the  centre  is  measured  by  its  intercepted  arc. 


PROPOSITION  TI. 

214,  Theorem. — An  angle  formed  ly  two  secants  meeting  zvith- 
out  the  circle  is  measured  hy  one-half  the  difference  of  the  intercepted 
arcs, 

Dem. — Let  APB  be  an  angle  formed  by  the  two  se- 
cants AP  and  PB  ;  then  is  it  measured  by  \  (AB  —  CD), 
i.  e.,  one-half  the  intercepted  arcs. 

For,  draw  CE  parallel  to  PB  ;  then  CD  =  EB  (i74), 
and  ACE  =  its  corresponding  angle  APB.  But  ACE  is 
measured  by  ^  AE=i  (AB-EB)=i  (AB-CD).    q.  e.  D. 


215,  ScH. — This  case  passes  into  that  of  an  in- 
scribed angle,  by  conceiving  P  to  move  toward  C,  thus 
diminishing  the  arc  CD.  When  P  reaches  C,  the  angle 
becomes  inscribed;  and,  as  CD  is  then  0,  i(AB  —  CD) 
=  ^AB.  Also,  by  conceiving  P  to  continue  to  move 
along  PA,  CD  will  reappear  oil  the  other  side  of  PA, 
hence  will  change  its  sign,*  and  ^(AE  —  CD)  will  become  i  (AE  +  CD),  as  it 
should,  since  the  angle  is  then  formed  by  two  chords  intersecting  within  the 
circumference. 


PROPOSITION  vn. 

216.  Theorem. — An  angle  formed  hy  a  tangent  and  a  chord 
draicn  from  the  point  of  tangency  is  measured  by  one-half  the  inter- 
cepted arc, 

*  In  accordance  with  the  law  of  positive  and  negative  quantities  as  used  in  mathematics, 
■whenever  a  continuously  varying  quantity  is  conceived  as  diminishing  till  it  reaches  0,  and 
then  as  reappearing  by  the  same  law  of  change,  it  must  change  its  sign. 


MEASUREMENT  OF  ANGLES. 


101 


Dem.— Let  TPA  be  an  angle  fonned  by  TM 
tangent  at  P,  and  the  chord  PA;  then  is  TPA 
measured  by  one-half  the  intercepted  arc  AP. 
For,  draw  any  chord  CD  parallel  to  TM  and 
cutting  AP.  Then  CEA  =  TPA.  But  CEA  is 
measured  by  i  (AC  +  PD).  Hence,  as  PD  =  CP 
(175),  TPA  is  measured  by  i  (AC  +  CP),  or  i 

A  P.      Q.  E.  D. 

Show  that  APM  is  measured  by  ^  arc  AmP. 
Also,  observe  how  the  case  of  two  secants 
{214)  passes  into  this. 


Fig.  161. 


PROPOSITION  TIU. 

217.  TJieorem, — An  angle  formed  iy  tivo  tangents  is  measured 
ly  one-lialf  the  difference  of  the  interce;pted 

arcs.  ,P 

Dem. — Let  APB  be  an  angle  formed  by  the 
two  tangents  AP  and  PB ;  then  is  it  measured  by 
i  {arc  CmD  —  arc  C?iD),  ^.  <?.,  one-half  the  differ- 
ence of  the  intercepted  arcs.  For,  through  one  of 
the  points  of  tangency,  as  C,  draw  a  chord,  as  CE, 
parallel  to  the  other  tangent.  Now,  ACE  is 
measured  by  i  arc  CE,  by  the  last  proposition. 
But  ACE  =  APB,  and  arc  CE  =  CmD  —  D??iE  = 
CwD  -  C;iD,  since  CnD  =  EmD  {175).  Hence, 
APB  is  measured  by  i  (CwD  —  CnD).    q.  e.  d. 

218,  SCH.— The  case  of  two  secants  {214) 
passes  into  this  by  supposing  the  secants  to 
separate  until  they  become  tangents. 


Fio.  162. 


EXERCISES. 

1.  JProb, — TJirough  a  given  point  to  draio  a  parallel  to  a  given 
line,  on  the  principles  contained  in  (149),  (201),  and  (1(>5). 

Solution.— Through  P  to  draw  a  parallel  to  AB.    Fiom  P  as  a  centre,  with 
any  radius  greater  than  the  distance 
from   P  to  AB,  describe  an  arc  cut-  . 

ting  AB,  asac.    From  a  as  a  centre,       ij\ 9<-= r^f N 

with  the  same  radius,  strike  an  arc  \ 

through   P,  intersecting    AB,  as   Ph. 

Take  the    chord    Pb    and    apply   it  ^5^ — ^ 

from  a  on  the  arc  ac,  as  aO.    These 

chords  being  equal,  the  arcs  Pb  and 


Fig.  163. 


102 


ELEMENTAKY  PLANE  GEOMETRY. 


Fitt.  164. 


rtO  are  equal  {165).  Again,  angle  Pdb  =  anglo 
OPa,  since  they  are  measured  by  equal  arcs  struck 
with  the  same  radius  (201).  These  alternate 
angles  being  equal,  MN  is  parallel  to  AB  {149). 

2.  In  Fig.  164  there  are  4  pairs  of  equal 
angles.  Which  are  they,  and  why  ?  Show 
also  that  COB  =  ABD  +  CDB,  by  (210), 
and  (212).  Show  also  that  DOB  =  ABC  + 
DAB. 


3.  I^rob, — From  a  point  without  a  circle  to  dratv  a  tangent  to 
the  circle. 

SoLUTiox. — Let  0  be  the  given  circle,  and  P  the  given  point.    Join  P  with 
the  centre  O,  and  upon  PO  as  a  diameter  describe  a  circle.     Let  T  and  T'  be 

the  intersections  of  the  two  cir- 
cumferences. Now,  if  lines  be 
drawn  from  P  through  T  and  T', 
they  will  be  tangent  to  the  cir' 
cle  O.  For  OTP  and  OT'P, 
being  inscribed  in  semi-circles, 
are  right  angles  {211).  Hence, 
PM  is  perpendicular  to  radius 
OT  at  its  extremity  T,  and 
is  therefore  a  tangent  {172). 
In  like  manner  PT'  is  shown 
to  be  a  tangent,  and  we  see 
that  from  a  point  without  a  cir- 
cle two  tangents  can  he  drawn  to 
ifie  circle. 


Fig.  165. 


4.  I^rob, — On  a  given  Ji?ie,  to  consti'uct  a  segment  ichich  shall 

contain  a  given  angle. 

SoLUTiox.  —  Let  AB  be 
the  given  line,  and  O  the 
given  angle.  At  one  ex- 
tremity of  the  given  line,  as 
B,  construct  an  angle  ABC 
equal  to  the  given  angle  O, 
which  shall  lie  on  the  oppo- 
site of  AB  from  that  on  which 
the  required  segment  is  to 
lie.  Erect  a  perpendicular 
to  the  line  CB  at  B,  and  also 
a  peq^endicular  bisecting 
AB.     Let    FB   and   FE    be 


MEASUREMENT  OF  ANGLES. 


103 


these  perpendiculars,  intersecting  at  F.  From  F  as  a  centre,  with  a  radius 
equal  to  FB,  describe  a  circle.  Then  is  AHB  the  segment  required.  For, 
CB  being  perpendicular  to  radius  FB  at  its  extremity,  is  tangent  to  the 
circle,  and  angle  ABC  (=  angle  0)  is  measured  by  ^  of  arc  AmB  {210).  Now, 
any  angle  -  inscribed  in  the  segment  km'm"B^  as  AHB,  has  ^  AwB  for  its 
measure,  and  is,  consequently,  equal  to  0. 

Another  Solution. — On  the  side  of  AB  on  w^hich  the  segment  is  to  lie,  draw 
any  line  through  either  extremity  of  AB, 
making  an  acute  angle  with  AB.  Let  CB 
be  such  a  line.  At  any  point  in  CB,  as 
C,  draw  a  line  CE,  making  angle  ECB  = 
the  given  angle  0,  Fig.  166.  Through  A 
pass  a  parallel  to  CE  (see  Ex.  1),  as  AD. 
Pass  a  circumference  through  A,  D,  and 
B.  Any  angle  inscribed  in  segment  AtwB 
is  equal  to  O.  [Let  the  student  give  all 
the  reasons,  and  make  the  construction. 
The  requisite  marks  for  the  construction 
are  made  in  the  figure.  Why  is  it  said, 
make  CBA  an  acute  angle?  When  would 
a  right  angle  answer?  When  an  obtuse 
angle?]  Fic.  167. 


SYNOPSIS. 


CO 

o 
< 
o 

H 
1^ 


How  angles  are  measured. 

r  Prop.  L  Equal  arcs  subtend  equal  (  g^^'  \  S['°Iure'of  1,  2,  and 
angles  at  the  centre.    ]  4  right  angles. 


O 


Prop.  II.  Commensurable  arcs  in  the  same  ratio 
as  their  subtended  angles. 


Cor.  Converse. 


Prop.  III.  Incommensurable  arcs. 


Cor.  Converse. 
iSch.  Method  of  measuring 
angles. 


Inscribed  angle,  what  ? 

Prop.  IV.  Inscribed  angle,  how  measured.  |  ^''^'-  "^'^g^emSh^le.^'  ^'^^  ^ 

Prop.  V.  Angle  between  two  chords.  ^Scli.  Compared  with  preceding. 

Prop.  VI.  Angle  between  two  secants.  [  Sch.  Compared  with  Prop.  IV. 

Prop.  VII.  Angle  between  tangent  and  chord. 

Prop.  VIII.  Angle  between  two  tangents.  \  Sch.  Compared  with  Prop.  VI. 

f  Prob.  To  draw  a  parallel  through  a  given  point. 

■p^..^nTci.«    1  P^'o^-  To  draw  a  tangent  to  a  circle  from  a  point  without 

j!.XERCisES.  <  p^.^j  rj.^  coustruct  a  segment  on  a  given  line  which  shall 

i.  contain  a  given  angle. 


104  ELEMENTARY  PLANE  GEOMETRY. 


SECTION    VII. 

OF  THE  ANGLES  OF  POLYGONS,  AND  THE  RELATION  BETWEEN 
THE  ANGLES  AND  SIDES. 


OF  TRIANGLES. 


PROPOSITION  I. 

%19,  Theorem, —  The  sum  of  the  three  angles  of  a  triangle  is 
tiuo  right  ajigles. 

Dem. — Conceive  a  circumference  passed  through 
the  vertices  of  the  triangle,  as  aitc,  through  the  ver- 
tices of  the  triangle  ABC  {58).  The  angle  A  is 
measured  by  \  arc  a,  B  by  i  h,  and  C  by  i  c.  Hence, 
A  +  B  +  C  is  measured  by  i  (a  +  6  +  c),  or  a 
semi-circumference,  and  is  equal  to  two  right  angles 
{2031    Q-  E.  D. 

Fig.  168.  220.  CoR.  1. — A  triangle  can  have  only 

m^e  right  angle,  or  one  oUuse  angle.    Why  ? 

221,  Cor.  2. — Two  angles  of  a  triangle,  or  their  su7n,  being 
given,  the  third  may  he  found  hy  subtracting  this  sum  from  two  right 
angles,  i.  e.,  either  angle  is  the  siqyplement  of  the  other  two. 

222,  Cor.  3. — The  sum  of  the  two  acute  aiigles  of  a  right-angled  tri- 
angle is  equal  to  one  right  angle;  i.e.,  they  are  comjjlements  of  each  other. 

223,  Cor.  4. — If  the  angles  of  a  triangle  are  equal  each  to  each, 
any  one  is  one-third  of  two  right  angles,  Oi  tico-thirds  of  one  right  angle. 


PROPOSITION  n. 
224,  TJieoreni, — The  sides  of  a  triangle  sustain  the  same 
GENERAL  relation  to  each  other  as  their  opposite  angles  ;  that  is,  the 
greatest  side  is  02)posite  the  greatest  angle,  the  second  greatest  side 
opposite  the  second  greatest  angle,  and  the  least  side  opposite  the  least 
angle. 


OF  THE  ANGLES  OF  TRIANGLES. 


105 


Fio 


Dem.— In  the  triangle  ABC  let  C  >  B  >  A  be 
the  order  of  the  values  of  the  angles;  then  AB  > 
AC  >  BC  is  the  order  of  the  values  of  the  sides. 

For,  circumscribe  the  circumference  dbc.  The 
angle  C  be-ing  greater  than  B,  the  arc  c,  the  half  of 
which  measures  C,  is  greater  than  the  arc  5,  the 
half  of  which  measures  B.  Now,  the  greater  arc 
has  the  greater  chord  {166).  Hence,  AB  >  AC. 
In  like  manner,  if  B  >  A,  arc  b  >  arc  a,  and  AC  > 
BC.  If  either  angle,  as  C,  is  obtuse,  AB  is  greater 
than  AC  or  BC,  because  it  lies  nearer  the  centre  {167)- 

225,  Cor.  1.— Conversely,  The  order  of  the  magnitudes  of  the 
sides  being  AB  >  AC>  BC,  the  order  of  the  magnitudes  of  the  angles  is 
C>  B  >  A. 

[Let  the  student  give  the  demonstration  in  form.] 

220,  Cor.  2. — An  equiangular  tria^igle  is 
also  equilateral ;  and,  conversely,  an  equilateral 
triangle  is  equiangular. 

Dem.— If  A  =  B  =  C,  arc  a  =  arc  h  —  arc  c,  and, 
consequently,  chord  BC  =  chord  AC  =  chord  AB. 
Conversely,  if  the  chords  are  equal,  the  arcs  are,  and 
hence  the  angles  subtended  by  these  arcs. 

227,  Cor.  3. — In  an  isosceles  triangle  the 
angles  opposite  the  equal  sides  are  equal; 
and,  conversely,  if  tioo  angles  of  a  triangle 
are  equal,  the  sides  opposite  are  equal,  and 
the  triangle  is  isosceles. 

Dem. — If  AB  =  BC,  arc  a  =  arc  c;  and  hence, 
angle  A,  measured  by  i  a,  =  angle  C,  measured  by 
i  c.  Conversely,  if  A  =  C.arca  —  arc  c ;  and  hence 
chord  BC  =  chord  AB. 

228,  ScH.— It  should  be  observed  that  the  proposition  gives  only  the  general 
relation  between  the  angles  and  sides  of  a  triangle. 

It  is  not  meant  that  the  sides  are  in  the  same  ratio 
as  their  opposite  angles  :  this  is  not  true.  Thus 
in  Fig.  172  angle  c  is  twice  as  great  as  angle  a ; 
but  side  c  is  not  twice  as  great  as  side  a,  although 
it  is  greater.  Trigonometry  discovers  the  exact 
relation  which  exists  between  the  sides  and 
angles. 


Fro.  172. 


106 


ELEMENTARY  PIA^'E   GEOMEITIY. 


PROPOSITION  m. 

229,    Tlieoreni, — If  from  any  j^oint  witJiui  a  triangle  lines 
be  draiun  to  the  extremities  of  any  sidcy  ths 

included  angle  is  greater  than  the  angle  of  the 
triangle  opposite  this  side. 

Dem. — Let  OA  and  OB  be  two  lines  drawn  from 
any  point  0  within  the  triangle  ABC,  to  the  extremi- 
ties of  the  side  AB  ;  then  angle  AOB  >  ACB. 
For,  circumscribe  a  circle  about  the  triangle.    Xow, 
,y  ACB  is  measured  by  \  AwB,  but  AOB  is  measured  by 

'^"'n^'  i(A7iB  +  EwiD).    Therefore,  AOB  >  ACB.     q.  e.  d. 

Fig.  173. 


230,  An  JExterior  Angle  of  a  polygou  is  an  angle  formed 
by  any  side  with  its  adjacent  side  produced,  as  CBD,  Fig.  174. 


PROPOSITION  IT. 

231,  TJieorem, — ^n  exterior  angle  of  a  triangle  is  equal  to  the 
sum  of  the  tivo  interior  non-adjacent  angUs. 

Dem.— Let  ABC  be  any  triangle,  and  CBD  an  ex- 
terior angle;  then  CBD  =  A  +  C. 

For  CBD  is  the  supplement  of  CBA  by  {131),  and 
CBA  is  the  supplement  of  A  +  C  by  {221).    Hence, 

CBD  rr  A    +    C.      Q.  E.  D. 

232.  Cor.— Either  angle  of  a  triangU  not 
adjacent  to  a  specified  exterior  angle,  is  equal 
to  the  difference  of  this  exterior  angle  and  the  other  non-adjacent 
angle. 

Thus,  since  CBD  =  A  +  0,  by  transposition,  CBD  —  A  =  C,  and  CBD  —  C 
=  A, 


OF  THE  ANGLES  OF   QUADRILATERALS. 


107 


OF   QUADRILATERALS. 


PROPOSITION  V. 

233,  Theorem, — The  sum  of  the  angles  of  a  quadrilateral  is 
four  right  angles. 

Dem. — Let  ABCD  be  any  quadrilateral ; 
then  DAB  +  ABC  +  BCD  +  CDA  =  four 
right  angles. 

For,  draw  either  diagonal,  as  AC,  di- 
viding the  quadrilateral  into  two  triangles. 
Then,  as  the  sum  of  the  angles  of  the  two 
triangles  is  the  same  as  the  sum  of  the  an- 
gles of  the  quadrilateral,  and  the  sum  of 
the  angles  of  the  triangles  is  twice  two 

right  angles  {219) ;  the  sum  of  the  angles  of  the  quadrilateral  is  four  right 
angles.    Q.  e.  d. 


PROPOSITION  TI. 

234,    Theorem. — The    oi^posite  angles  of  any  quadrilateral 

which  can  he  inscribed  in  a  circle  are  supj^lementary. 

Dem. — Let  ABCD  be  any  inscribed  quadrilateral; 
then  A  -f  C  =  two  right  angles^  and  D  -f  B  =  two 
right  angles. 

For,  A  is  measured  by  |  arc  BCD,  and  C  is  meas- 
ured by  i  arc  DAB  (210).  Hence,  A  +  C  is  meas- 
ured by  one-half  a  circumference,  and  is,  therefore, 
equal  to  two  right  angles  {203).  In  like  manner  D 
is  measured  by  i  arc  ABC,  and  B  by  i  arc  ADC. 
Consequently,  D  +  B  is  measured  by  one-half  a  cir-' 
cumference,  and  is,  therefore,  equal  to  two  right 
angles. 


PROPOSITION  m, 

23S,  Hieorem, — The  opposite  angles  of  a  parallelogram  are 
equal,  and  the  adjacent  angles  are  supplementary. 

Dem.— ABCD,  Fig.  177,  being  any  parallelogram,  A  =  C,  B  r=  D,  and  B  -h  C, 
C  -f-  D,  D  4-  A,  and  A  +  B,  each  =  two  rigid  angles. 


108  ELEMENTARY  PLANE  GEOMETRY. 

For,  produce  any  side,  as  AB,  fonn- 

^ ing  the  two  exterior  angles  EAD  and 

y/^  CBF.     Since   CB   and  DA  are  parallel, 

/  and  FE  cuts  them,  the  corresponding  an- 

^ P  gles,   CBF   and  DAB   are  equal  {152). 

'^  FiQ.  177.  Again,  since  EF  and  DC  are  parallel,  and 

CB  cuts  them,  the  alternate  interior* 
cngles  CBF  and  C  are  equal  {152).  Hence,  as  DAB  and  C  are  each  equal  to 
CBF,  they  are  equal  to  each  other.  Li  a  similar  manner  D  can  be  proved  equal 
to  CBA.     [Let  the  student  give  the  proof.] 

That  the  angles  B  and  C  of  the  parallelogram  are  supplemental  is  evident 
from  {150),  which  proves  that  the  sum  of  two  interior  angles  on  the  same  side 
of  a  secant  cutting  two  parallels  is  two  right  angles.  For  a  like  reason  A  -i-  D 
=  ttco  right  angles^  etc. 

236.  CoK.  1. — Tlie  ttvo  angles  of  a  trape- 
/  \         zoid  adjacent  to  either  one  of  the  two  sides 

/  \       not  parallel  are  supplemental. 


Fig.  its.  [Let  the  student  show  why.] 

237.  Cor.  2. — If  one  angle  of  a  parallelogram  is  right,  the  others 
are  also,  and  the  figure  is  a  rectangle. 


PROPOSITION  Tin. 

238,  TJieoreni, — Conversely  to  the  last,  If  the  adjacent  angles 
of  a  quadrilateral  are  supplementary,  or  the  opposite  angles  equal, 
the  figure  is  a  parallelogram. 

Dek.— If  A  +  D  =  two  right  angles,  AB  and  DC  are  parallel  by  {147). 

For  a  like  reason,  if  D  +  C  =  two  right 
angles,  DA  and  CB  are  parallel.  Again, 
if  A  =  C  and  D  =  B,  by  adding  we 
have  A  +  D  =  C+B.  ButA  +  D  + 
C  +  B  =r  four  right  angles  {233). 
Hence,  A  +  D  =  two  right  angles,  and 
^'''-  ^'^-  AB  and  CD  are  parallel.     So,  also,  A  + 

B  can  be  shown  to  be  equal  to  two  right  angles ;  and,  consequently,  AD  and 

CB  are  parallel. 


*  Interior  with  reference  to  the  parallels  (146). 


OF   THE   SIDES   OF   QUADRILATERALS. 


109 


PROPOSITION  EX.      , 

239*  Theorem. — If  two  op])osite  sides  of  a  quadrilateral  ar^ 
equal  and  parallel,  the  figure  is  a  2:>arallelogram. 

Dem.— In  (a)  let  DC  be  equal 
and  parallel  to  AB  ;  then  is  ABCD 
a  parallelogram. 

For,  drawing  the  diagonal 
AC,  it  makes  the  angles  ACD  and 
CAB  equal,  since  they  are  altern- 
ate intenor  angles  (15^).  Con- 
ceive the  quadrilateral  divided 
in  this  diagonal  into  two  tri- 
angles, as  in  (b).  Reverse  the 
triangle  ACB  and  place  it  as  in 
(c).  Draw  DB.  Since  angle  DCA 
=  angle  CAB,  and  DC  =  BA,  if 
CBA  be  revolved  upon  AC,  AB 
will  take  the  direction  CD,  B  will 
fall  in  D,  and  CBA  will  coin- 
cide with  ADC.  Hence,  angle 
ACB  =  angle  DAC.  But  in  (a) 
these  are  alternate  interior  an- 
gles made  by  AC  with  AD  and  BC 

Q.   E.   D. 


Therefore,  AD  is  parallel  to  BC  (149). 


PROPOSITION  X. 

240,  Tlieoveni, — If  the  opjjosite  sides  of  a  quadrilateral  are 
equal,  the  figure  is  a  parallelogr'am. 

Dem.— In  (a)  let  AB  =  DC,  D  C 

and  AD  =  BC ;  then  is  ABCD  a 
parallelogram. 

For,  divide  the  quadrilateral 
in  the  diagonal  AC,  and  revers- 
ing the  triangle  ABC,  place  it 
as  in  (c),  and  draw  DB.  Since 
AB  ^  CD,  and  CB  =  AD,  DB  is 
perpendicular  to  CA  {130). 
Now,  revolving  ABC  upon  CA, 
it  will  coincide  with  ADC.  Hence, 
angle  DCA  =  angle  CAB,  and 
AB  is  parallel  to  DC.  Also, 
angle  DAC  =  angle  BCA,  and 
AD  is  parallel  to  BC.  There- 
fore, ABCD  is  a  parallelogram. 

Q.   E.   D. 


Fio.  181. 


110 


ELEMENTARY   PLANE  GEOMETRY. 


PROPOSITION  XI. 

241.  TJieorem, — Conversely  to  the  last,  Tlie  opposite  sides  of 
a  parallelogram  are  equal. 

D C  Dem.— Let  ABCD  be  a  paral- 

lelo<rram ;  then  AB  =  DC,  and 
AD  =  CB. 

Since  DC  is  parallel  to  A B,  an- 
gle DCA=  angle  CAB,  Also,  since 
AD  is  parallel  to  BC,  angle  DAC 
=  angle  ACB  {132).  Now,  divide 
the  parallelogram  (a)  in  the  di- 
agonal, and  place  ABC  as  in  (c). 
Revolve  ABC  on  AC,  until  it  falls 
in  the  plane  on  the  other  side  of 

AC.  Since  angle  BAC  =  angle 
ACD,  AB  will  take  the  direction 
CD,  and  B  will  fall  in  CD,  or  CD 
produced.  Since  angle  BCA  = 
angle  DAC,  CB  will  take  the 
direction  AD,  and  B  will  foil  in 

AD,  or  in  AD  produced.  There- 
fore, as  B'falls  at  the  same  time  in  AD  and  CD,  it  falls  at  the  intersection  D,  and 
the  triangles  coincide.     Hence,  AB  =  CD,  and  AD  =  CB.     q.  e.  d. 

242.  Cor.  1. — Parallels  intercepted  letioeen  parallels  are  equal. 

243.  Cor.  2. — A  diagoiial  of  a  parallelogram  divides  it  into  tic o 
equal  triangles. 


Fig.  1S2. 


244.  TJteoreni. 

oisect  each  other. 


PROPOSITION  XII. 

The  diagonals  of  a  parallelogram  mutually 


Dem.— Let  AC  and  DB  be  the  diagonals  of 
the  parallelogram  ABCD  («),  and  Q  their  inter- 
section ;  then,  DQ  =  QB,  and  AQ  =  QC. 

For,    take    the    triangle  AQB,  and  apply  it  to 

DQ'C,  by  placing  BA  in  its  equal  DC,  B  falling  at 

D,  and  A  at  C,  with  the  vertices  Q  and  Q'  on  the 

same  side  of  this  common  line,  as  in  (6).    Now, 

since  angle  QBA  =  Q'DC  {152\  BQ  will  take  the 

direction  DQ',  and  Q  will  fall  in  DQ',  or  in  DQ' 

Fig.  183.  produced.     For  a  like  reason    AQ  will   take  the 

direction  CQ',  and  Q  will  foil  in  CQ',  or  in  CQ'  produced.    Hence,  as  Q  falls 

at  the  same  time  in  DQ'  and  CQ',  it  falls  at  their  intersection  Q' ;  whence 

BQ  =  DQ',  and  AQ  =  CQ'.     Q.  e.  d. 


D 

c 

/■"••■■■ 

•--q:--" 

--"/ 

/ 

/ 

-"■Q'->. 

/(^) 

/ 

-__/ 

I-\ 

B 

A^ 

q;.- 


(«») 


OF   THE   DIAGONALS   OF   QUADRILATERALS. 


Ill 


PROPOSITION   XHL 

245,  TJieorem. — The  diagojials  of  a  rJiomhus  bisect  each  other 
at  right  angles. 

Dem. — Let  AC  and  DB  be  the  diagonals  of  the 
rhombus  ABCD ;  then  are  they  at  right  angles  to 
each  other,  and  bisected  at  Q. 

For,  since  AB  =  AD,  and  DC  =  CB,  AC  has  two 
of  its  points  equally  distant  from  D  and  B,  and  is, 
therefore,  perpendicular  to  DB,  at  its  middle  point 
{130).  In  like  manner  D  and  B  are  each  equidistant 
from  A  and  C,  whence  Q  is  the  middle  point  of  AC. 


.-"'Q- 


A 


Fig.  184. 


-/C 


246.  Cor. — The  diagonals  of  a  rhomhus  lisect  its  angles. 

For,  revolve  ABC  upon  AC  as  an  axis,  and  it  will  coincide  with  ADC.  Hence 
angles  A  and  C  are  bisected.'  In  like  manner  revolve  DAB  upon  DB,  and  it  will 
coincide  with  DCB.    Hence  D  and  B  are  bisected. 


247.  Theorem.- 

gle  are  equal. 


PROPOSITION  XIV. 

Tlie  diagonals  of  a  rectan- 


Dem.— Let  AC  and  DB  be  the  diagonals  of  the  rectan- 
gle ABCD  ;  then  AC  =  DB. 

For,  upon  AC  as  a  diameter  describe  a  circle.  Since 
D  and  B  are  right  angles,  they  are  inscribed  in  semicir- 
cles {211),  and  DB  is  a  diameter.    Therefore,  AC  =  DB. 

Q.  E.  D. 


Fig.  185. 


248.  CoR. — Conversely,  If  the  diagonals  of  a  parallelogram  arc 
equal,  the  figure  is  a  rectangle. 

Dem.— Since  the  diagonals  of  a  parallelogram  bisect  each  other,  if  they  are 
equal,  a  circumference  described  from  their  intersection  as  a  centre,  with  a 
radius  equal  to  half  of  a  diagonal,  will  pass  through  the  vertices  of  the  parallel- 
ogram. Hence  the  diagonals  will  be  diameters,  and  the  angles  will  be  inscribed 
in  semicircles,  and  consequently  will  be  right  angles. 


112 


ELEMENTARY  PLANE  GEOMETRY. 


OF  POLYGONS  OF  MORE  THAX  FOUR  SIDES. 

240,  A  Salient  Anr/le  of  a  polygon  is  one  whose  sides,  when 
produced,  can  only  extend  wifJiout  the  polygon. 

250,  A  JRe'entrant  Anr/le  of  a 

polygon  is  one  whose  sides,  when  pro- 
duced, can  extend  with  171  the  polygon. 

III.— In  the  polygon  ABCDEFC,  all  the  an- 
gles are  salient  except  D,  which  is  re-entrant. 

251.  A  Convex  JPolyf/ofi    is    a 

polygon  which  has  only  salient  angles. 
A  polygon  is  always  supposed  to  be  con- 
vex, unless  the  contrary  is  stated. 

A  Concave  or  Re-entrant  JPolygon  is  a  polygon 
with  at  least  one  re-entrant  angle. 


0x0 


PROPOSITION   XT. 

253,  TJieoreni, —  The  sum  of  the  interior  angles  of  a  polygon 
is  equal  to  tiuice  as  many  right  angles  as  the  polygon  has  sides,  less 


four  right  angles. 

F 


Dem. — Let  n  be  the  number  of  sides  of  any 
polygon ;  then  the  sum  of  its  angles  is 

n  times  two  rigTit  angles  —  4  right  angles. 

For,  from  any  point  O,  will) in,  draw  lines  to 
the  vertices  of  the  angles.  As  many  triangles 
will  thus  be  formed  as  the  polygon  has  sides,  that 
is,  n.    The  sum  of  the  angles  of  these  triangles  ia 

n  times  two  Hght  angles  {219). 

But  this  exceeds  the  sum  of  the  angles  of  the 
polygon  by  the  sum  of  the  angles  at  the  common  vertex  0,  that  is,  by  4  right 
angles.     Hence  the  sum  of  the  angles  of  the  polygon  is 

n  times  two  right  angles  —  4  right  angles,     q.  e.  d. 

25^,  ScH.  1. — The  sum  of  the  angles  of  a  pentagon  is  5  times  tico  right  an- 
gles —  4  right  angles^  or  6  right  angles.  The  sum  of  the  angles  of  a  hexagon  is 
8  right  angles;  of  a  heptagon,  10 ;  of  an  octagon,  12,  etc. 


Fig.  187. 


OF  THE  ANGLES  OF  REGULAR  POLYGONS. 


113 


255,  ScH.  2.— This  proposition  is  equally  applicable  to  triangles  and  to 
quadrilaterals.  Tlius  the  sum  of  the  angles  of  a  triangle  is  3  times  two  right 
angles  —  4  nght  angles  (or  6  —  4)  =  2  right  angles.  So  also  the  sum  of  the 
angles  of  a  quadrilateral  is  4  times  tico  Tight  angles  —  4  right  angles,  or  4  right 
angles. 

256,  ScH.  3. — To  find  the  value  of  an  angle  of  an  equiangular  polygon, 
that  is,  one  whose  angles  are  equal  each  to  each,  divide  the  sum  of  all  the 
angles  by  the  number  of  angles. 


PROPOSITION  XYI. 


m:.^ 


257.  Theore'in, — If  the  sides  of  a  polygon  he  produced  so  as  to 
form  one  exterior  angle  {and  only  one)  at  each  vertex,  the  snm  of 
these  exterior  angles  is  four  rigid  angles. 

Dem. — Let  n  be  the  number  of  sides  of  ahy 
polygon.  At  each  of  the  n  angles,  there  is  an 
interior  and  an  exterior  angle,  whose  sum,  as 
A  +  a,  is  two  right  angles.  Hence  the  sum  of 
all  the  exterior  and  interior  angles  is  n  times  two 
right  angles.  Now,  from  this  sum  subtracting 
the  sum  of  the  exterior  angles,  the  remainder 
is  the  sum  of  the  interior  angles.  But,  by  tlie 
preceding  proposition,  4  nglit  angles  subtracted 
from  n  times  tico  right  angles,  leaves  the  sum 
of  the  interior  angles.  Therefore  the  sum  of 
the  exterior  angles  is  4  right  angles,    q.  e.  d.  Fig.  18S. 


OF  REGULAR  POLYGONS* 


PROPOSITION  xrn. 

258.  Tlieorem. — The  angles  of  an  inscribed  equilateral  polygon 
are  equal ;  and  the  polygon  is  regular, 

Dem.— Let  ABCDEF  be  an  inscribed  polygon,  with  AB  =  BC  =  CD,etc.: 
then  is  angle  A  =  B  =  C  =  D,  etc.,  and  the  polygon  is  regular. 

For, from  the  centre  of  the  circle  draw  OF,  OA,  and  OB,  and  also  the  per- 
pendiculars Oa  and  06.    Revolve  OFA  upon  OA  as  an  axis,  until  it  falls  in  the 

8 


114 


ELEMENTARY   PLANE   GEOAIETRY. 


plane  of  OAB. 
D 


Since  the  chords  FA  and  AB  are  equal,  the  arc  FA  =  arc  AB, 
and  F  falls  at  B.  Hence  the  triangles  OFA  and  OAB 
coincide.  The  angle  A  of  the  ])oiygon  is  therefore 
bisected  by  OA ;  that  is,  OAF  =  OAB.  In  the  same 
manner  OBA  can  be  shown  equal  to  OBC.  More- 
over, since  O  A  and  OB  are  equal  oblique  lines  drawn 
from  a  point  in  the  perpendicular,  angle  OAB  = 
angle  OBA.  Hence,  as  the  half  of  A  equals  the  half 
of  B,  A  =  B.  In  like  manner,  B  can  be  shown  equal 
to  C,  C  to  D,  D  to  E,  etc.  Therefore  the  polygon  is 
equiangular,  as  well  as  equilateral,  and  consequently 
regular  {117).    Q.  e.  d. 


PROPOSITION  XTin. 

2o0»  TJieovem. — T7ie  sides  of  an  inscribed  eqxdanguJar polygon 
are  equal  wlien  their  number  is  odd  ;  and  the  polygon  is  regular. 

Dem. — Let  ABCDEFG  be  an  inscribed  equiangular  polygon  of  an  odd 
number  of  sides;  then  is  side  AB  =  BC  =  CD,  etc., 
and  the  polygon  is  regular. 

For,   from  the  centime  of  the  circle  draw  the 
radii  OA,  OB,  etc.,  to  tlie  vertices  of  the  polygon, 
and  Oa,  Ob,  etc.,  perpendicular  to  the  sides.     Re- 
volve the  quadrilateral  GGA«,  upon  Oa  as  an  axis 
until  it  falls  in  the  plane  of  OCBa.      Since   0^ 
is   perpendicular  to  the  chord  AB,  ^a  =  aB,  and 
A  will  f;\ll  at  B.     Also,  as  the  angle  A  of  the  poly- 
gon =  B,  AG  will  fall  in  BC.    Now  G  falls  at  the 
same  time  in  the  arc  BCD  {138)  and  in  BC,  and 
hence  falls  at  their  intersection  C.     Therefore  AG 
=  BC.    In  like  manner  revolving  OBCc  upon  Oc 
as  an  axis,  BC  is  found  equal  to  ED.     So  also  we 
can  show  that  ED  =  FG  ;  then  that  FG  =  AB  ;  then  that  AB  =  DC  ;  and  finally, 
that  DC  =  EF.      Hence  we  have  GA  =  BC  =  ED  =  FG  =  AB  =  DC  =  EF  ;  and 
as  the  polygon  is  equiangular  by  hypothesis,  it  is  regular  {117)-    Q-  e.  d. 

260,  ScH.— It  is  easy  to  see  that  the  above  argu- 
ment would  fail  in  the  case  of  a  polygon  of  an  even 
number  of  sides,  because,  in  going  around  the  second 
time  the  same  sides  would  coincide  as  in  going  around 
the  first  time.  Moreover,  we  can  readily  inscribe  an 
equiangular  polygon  of  an  even  number  of  sides  which 

shall  not  be  regular. 
Fie.  191. 


Fig.  190. 


OF  THE   SIDES   AND   ANGLES   OF  REGULAR   POLYGONS. 


115 


PROPOSITION  XIX. 


< 


2S1*  Theovein, — TJie  sides  of  a  circiunscrihed  equiangular 
polygon  are  equal ;  and  the  polygon  is  regular. 

Dem. — Let  ABCDEF  be  a  circumscribed  polygon,  with  angle  A  =  B  =  C, 
etc. ;  then  is  AB  =  BC  =  CD,  etc.,    and  the  polygon  is  regular. 

For,  from  the  centre  of  the  circle,  draw  OA,  OB, 
etc.,  to  the  vertices  of  the  polygon,  and  Oa,  06,  etc., 
to  the  points  of  tangency.  The  latter  will  be  per- 
pendicular to  the  sides  by  {173).  Now  reverfe  the 
triangle  AaO,  and  apply  it  to  A60,  placing  Oa  in  its 
equal  06  ;  aA  will  take  the  direction  6A.  Then  will 
OA  of  the  triangle  AaO,  fall  in  OA  of  the  triangle 
A&O,  since  there  cannot  be  two  equal  oblique  lines  on 
the  same  side  of  Oh  {140).  Hence  angle  bAO  =  angle 
aAO,  and  bA  =  ak.    In  the  same  way  it  can  be  Fig.  192. 

shown  that  OB,  OF,  etc.,  bisect  the  other  angles,  and  that  bS  =  Be,  etc. 
Whence,  as  the  polygon  is  equiangular,  these  halves  are  equal,  that  is,  OAa 
=  OFa,  etc.  Then,  as  OA  and  OF  make  equal  angles  with  AF,  they  cut  off 
equal  distances  from  a,  and  Aa  —  aF.  So,  likewise,  we  can  show  that  Ab  =  6B, 
and  that  each  side  is  bisected  at  the  point  of  tangency.  Therefore,  as  the  halves 
of  the  sides  are  equal,  the  polygon  is  equilateral,  as  well  as  equiangular,  and 
consequently  regular  {117).    Q.  e.  d.'- 


PROPOSITION   XX. 

262,  Hieorem, — The  angles  of  a  circumscribed  equilateral 
polygon  are  equal  when  their  number  is  odd  ;  and  the  polygon  is 
regular. 

Dem. — Let  ABODE  be  a  circumscribed  polygon 
with  AB  =  BC  =  CD,  etc. ;  then  is  angle  A  =  3 
=  C  =  D,  etc.,   and  the  polygon  is  regular. 

In  the  same  manner  as  in  the  preceding  demon- 
stration, we  may  show  that  OA,  OB,  etc.,  bisect 
the  angles  of  the  polygon.  [The  student  should 
go  through  the  process.]  Then  revolve  the  tri- 
angle AOE  upon  AO  as  an  axis  till  it  falls  in  the 
plane  of  AOB ;  and  as  angle  OAE  =  angle  OAB, 
and  AE  =:  AB,  the  triangles  will  coincide.  Hence 
angle  OEA,  the  half  of  angle  E  of  the  polygon.  Fig.  in.3. 

equals  angle  OBA  the  half  of  B,  and  E  =  B.  In  like  manner  revolving  AOB 
upon  OB,  we  can  show  that  A  =  C.  So  also  we  find  B  =  D,  and  D  =  A. 
Therefore  the  polygon  is  equiangular  as  well  as  equilateral,  and  consequently 
regular,    q.  e.  d. 


116 


ELEMENTAEY  PLANE  GEOMETRY. 


263,  ScH.— That  the  above  style  of  argument  fails  in  the  case  of  a  polygon  of 
an  eveji  number  of  sides,  may  be  observed  by  attempt- 
ing to  apply  it.  Thus,  from  Fig.  192,  we  would  have 
A  =  C,  B  =  D,  C  =  E,  D  =r  F,  E  =  A,  and  F  =  B. 
From  these  we  have  A  =  C  =:  E,  and  B  =  D  =  F. 
But  the  process  will  not  give  any  one  of  the  first  three 
Fig.  194.  angles  equal  to  any  one  of  the  second  set.     That  is, 

it  Joes  not  follow  that  two  adjacent  angles  are  equal  in  case  the  number  of  sides 
is  even.  We  can  readily  construct  a  circumscribed  equilateral  polygon  which 
shall  not  be  equiangular. 


PROPOSITION    XXI. 

264,  TJieorem, — A  circumference  may  he  circumscribed  about 
any  regular  'polygon, 

Dem. — Let  ABCDEFbe  a  regular  polygon.  Bisect  A F  with  a  perpendicular 
Oa.  Any  point  in  this  perpendicular  is  equidistant 
from  A  and  F.  Bisect  AB,  adjacent  to  AF,  with  a 
perpendicular,  as  Oh.  Any  point  in  this  perpendic- 
ular is  equidistant  from  A  and  B.  Hence  the  inter- 
section of  these  perpendiculars,  O,  is  equidistant  from 

A,  F,  and  B,  and  a  circumference  described  from  0  as 
a  centre,  with  a  radius  OA,  will  pass  through  F  and 

B.  Now  revolve  the  quadrilateral  FOM  upon  Oh  as 
an  axis  until  it  falls  in  the  plane  of  COJB,  hk  will 
fall  in  its  equal  6B  ;  and  since  angle  A  =  angle  B, 
and  side  AF  =  side  BC,  F  will  fall  at  C.  Thus  it 
appears  that  the  circumference  described  from  0, 
and  passing  through  F,  A,  and  B,  also  passes  through 

C.     In  a  similar  manner  it  can  be  shown  that  the  same  cu-cumfcrence  passes 
through  all  the  vertices,  and  hence  is  circumscribed.      Q.  e.  d. 

265,  Cor.  1. — A  circumference  may  be  inscribed  in  any  regular 
polygon. 

Dem.— For,  having  circumscribed  one  about  it,  the  equal  sides  become  equal 
chords,  aud  hence  are  equally  distant  from  the  centre.  If,  therefore,  a  circle  be 
drawn  from  0  as  a  centre,  with  Oa  as  a  radius,  it  will  touch  every  side  of  the 
polygon  at  its  middle  point. 

266,  Cor.  2. — Tlie  ceiitres  of  the  inscribed  and  circumscribed 
circles  coincide. 

267,  Tlie  Centre  of  a  regular  polygon  is  the  common  centre 
of  its  inscribed  and  circumscribed  circles. 


OF   THE   SIDES   OF   POLYGONS. 


117 


268.  An  Angle  at  the  Centre  of  a  regular  polygon  is  the 
angle  included  by  two  lines  drawn  from  the  centre  to  the  extremities 
of  a  side,  as  FOA,  AOB. 

269,  Co^.  3. — The  angles  at  the  centre  of  a  regular  polygon  are 
equal  each  to  each;  and  any  one  is  equal  to  four  right  angles  divided 
by  the  number  of  sides  of  the  polygon. 

270*  The  Apothem  of  a  regular  polygon  is  the  distance  from 
the  centre  to  any  side,  and  is  the  radius  of  the  inscribed  circle. 


PROPOSITION  XXII. 

271.  Theorem. — The  side  of  a  regular  inscribed  hexagon  is 
equal  to  the  radius. 

Dem. — Let  ABCDEF  be  a  regular  inscribed  hexagon ;  then  is  any  side,  as 
BC,  equal  to  OB,  the  radius. 

In  the  triangle  BOC  the  angle  O  is  measured  by 
the  arc  BC,  or  i  of  a  circumfejence,  and  hence  is  \ 
of  4  right  angles,  or  |  of  a  right  angle.  Angle  ABC 
is  measured  by  ^  arc  CDEFA,  or  f  of  a  circumfer- 
ence. Hence  angle  OBC,  which  is  ^  of  ABC,  is 
measured  by  ^  of  f ,  or  |  of  a  circumference,  and  is, 
consequently,  equal  to  BOC.  So  also  OCB,  the  half 
of  DCB,  is  measured  by  ^  of  a  circumference.  Hence 
OCB  is  equiangular,  and  consequently  equilateral 
{226),  and  BC  =  OB.    q.  e.  d. 


272.  A  Broken  Line  is  said  to  be  Convex  when  no  one  of  its 
parts  will,  when  produced,  enter  the  space  included  between  it  and 
a  line  joining  its  extremities. 


PROPOSITION  xxin. 

273.  Theore'kn. — A  Convex  broken  line  is  less  than  any  broken 
line  luhich  envelops  it  and  has  the  same  extremities. 

Dem. — Let  khcdB  be  a  broken  line  enveloped 
by  the  broken  line  ACDEFB,  and  having  the 
same  extremities  A  and  B ;  then  is  KbcdS  < 
ACDEFB. 

For,  produce  the  parts  of  ^hcdB  till  they  meet 
the  enveloping  line,  as  A6  to  e,  be  to  /,  and  cd 
to  g.  Now,  since  a  straight  line  is  the  shortest 
path  between  two  points,  Ae  <  ACe,  bf  <  beDEf, 


Fig.  197. 


118 


ELEMENTARY  PLANE  GEOMETRY. 


^9  <  rf^9,  and  dB  <  dgS.  Hence,  if  a  point  starts  from  A  to  move  to  B,  AeDEFB 
will  be  a  shorter  path  than  ACDEFB,  A6/FB  shorter  than  A^DEFB,  KhcgB  shorter 
than  A5/FB,  and  AJcrfB  shorter  than  khcgB.  Therefore,  kbcdB  is  shorter  than 
ACDEFB.     Q.  E.  D. 

274,  Cor.  1. — The  sum  of  any  two  sides  of  a  triangle  is  greater 
than  the  third  side. 

This  is  the  same  as  the  axiom  that  the  shortest  distance  between  two  points 
is  a  straight  line. 


275, 


Cor.  2. — The  difference  hetween  any  tivo  sides  of  a  tria^igle 
is  less  than  the  third  side. 

Dem. — Let  rt,  6,  and  c  be  the  sides.  By  Corollary  1st,  a  +  &  >  c.  Therefore, 
transposing,  a>  c  —  b. 

276,  Cor.  3. — If  from  any  point  ivithin  a  tria^igle  liiies  he 
draton  to  the  extremities  of  any  side,  the  sum  of  these  lines  is  less 
than  the  sum  of  the  other  two  sides  of  the  triangle. 


Fio.  198. 


EXERCISES. 

1.  Giyen  two  angles  of  a  triangle,  to 
find  the  third. 

Bug's. — The  student  should  draw  two  angles 
on  the  blackboard,  as  a  and  b,  and  then  proceed 
to  find  the  thhd.  The  figure  will  suggest  the 
method.     The  third  angle  is  c. 

The  solution  is  effected  also  by  constructing 
the  two  given  angles  at  the  extremities  of  any 
line,  and  producing  the  sides  till  they  meet. 


2.  Two  angles  of  a  triangle  are  re- 
spectively f  and  1^  of  a  right  angle.     What  is  the  third  angle  ? 

3.  The  angles  of  a  triangle  are  respectively  |,  J,  and  f  of  a  right 
angle.  Which  is  the  greatest  side  ?  Which  the  least  ?  Can  you  tell 
the  ratio  of  the  sides  ? 

4.  What  is  the  value  of  one  of  the  equal  angles  of  an  isosceles  tri- 
angle whose  third  angle  is  ^  of  a  right  angle  ? 

5.  Two  consecutive  angles  of  a  quadrilateral  are  respectively  f  and 
I  of  a  right  angle,  and  the  other  two  angles  are  mutually  equal  to 


OF  THE   SIDES  AND  ANGLES   OF   POLYGONS.  119 

each  other.    What  is  the  form  of  the  quadrilateral  ?    What  the  value 
of  each  of  the  two  latter  angles  ? 

6.  One  of  the  angles  of  a  parallelogram  is  f  of  a  right  angle.  What 
are  the  values  of  the  other  angles  ? 

7.  The  two  opposite  angles  of  a  quadrilateral  are  respectively  -I 
and  f  of  a  right  angle.  Can  a  circumference  be  circumscribed  ?  If 
so,  do  it. 

8.  Two  of  the  opposite  sides  of  a  quadrilateral  are  parallel,  and 
each  is  15  in  length.  What  is  the  figure  ?  Do  these  facts  determine 
the  angles  ? 

9.  Two  of  the  opposite  sides  of  a  quadrilateral  are  12  each,  and  the 
other'two  7  each.  What  do  these  facts  determine  with  reference  to 
the  form  of  the  figure  ? 

10.  What  is  the  value  of  an  angle  of  a  regular  dodecagon  ? 

11.  What  is  the  sum  of  the  angles  of  a  nonagon  ?  What  is  the 
value  of  one  angle  of  a  regular  nonagon  ?     Of  one  exterior  angle  ? 

12.  What  is  the^  regular  polygon,  one  of  whose  angles  is  l|f  right 
angles  ?       /^/^^J^-^ 

13.  What  is  the  regular  polygon,  one  of  whose  exterior  angles 
is  f  of  a  right  angle  ? 

14.  Can  you  cover  a  plane  surface  with  equilateral  triangles  with- 
out overlapping  them  or  leaving  vacant  spaces?  With  quadrilat- 
erals? Of  what  form?  With  pentagons?  Why?  With  hexagons? 
Why  ?  What  insect  puts  the  latter  fact  to  practical  use  ?  Can  you 
cover  a  plane  surface  thus  with  regular  polygons  of  more  than  6 
sides  ?    Why  ? 

15.  Is  an  equilateral  hexagon  circumscribed  about  a  circle  neces- 
sarily regular  ?     A  heptagon  ?     An  octagon  ?     A  nonagon  ? 

16.  Is  an  equiangular  circumscribed  quadrilateral  necessarily  reg- 
ular ?     A  pentagon  ?     A  hexagon  ?     A  heptagon  ? 

17.  Is  an  equilateral  inscribed  pentagon  necessarily  regular  ?  An 
octagon?  How  is  it  if  they  are  equiangular;  are  they  necessarily 
equilateral  and  regular  ? 


120 


ELEMENTAEY   PLANE   GEOMETRY. 


SYNOPSIS. 


CO 

o 

c 

o 

'O 

m 

Q 
»2 


c 

9 

5zi 

Q 

-< 

C» 

cc 

'"^ 

(=J 

^ 

t^ 

<: 

c 

>^; 

HZ 

•< 

S 

p; 

o 

S 

< 


Prop. 


L  Sum  of  angles. 


f  Cor.  1. 

I  a>r.  2. 

1  Cor.  3. 

I  (7(>r.  4. 


Only  one  right  or  obtuse. 
Two  angles  given. 
Acute  angles  if  right  angled 
One  angle  if  equiangular. 


Cor 
Cor. 


Prop.     II.  Sides  and  opp.  angles.  ■{   Cor. 

I  ScJi. 
I 


Converse. 

2.  Equiangular,    equilat- 

eral, and  converse. 

3.  Isosceles,  equiangular, 

and  converse. 
These  only  general  rela- 
tions. 


Prop.  Ill,  Angle  within  a  triangle. 
Def.  Exterior  angle. 

I  Prop.   IV.  Exterior  angle.— C<?r.  Non-adjacent  interior. 


f  Prop. 
Prop. 


y.  Sum  of  angles. 
VI.  Angles  of  inscribed. 


i  s 

o 


Prop.    VII 
VIII 


Angles  of. 


\  Cor. 


Of  a  trapezoid. 
Of  a  rectansrle. 


"^    I 


{  Cor. 
Converse  to  last. 
IX.  Two  op.  sides  of  a  quadrilat'l  equal  and  parallel. 
X.  Opposite  sides  of  a  quadrilateral  equal.  tparaUelg. 
XI.   Convei-se  to  last.  ^  C;ar.   l.  Parallels  intercepted  bet. 
\  Cor.  2.  Diagonal  of  a  parallelogram. 
rPROP.    XII.  Bisect. 
Diagonals.  \  Prop.  XIIL  Of  a  rhombus.— C5?r.  Bisect  angles. 
(Prop.  XIV.  Of  a  rectangle. — Cor.  Converse. 


Prop. 
Prop. 
Prop. 

Prop. 


Prop.    XV.  Sum  of  ansrles. 


«  \ 


'  Def's. — Salient  angle. — Re-entrant. — Convex  polygon. — Concave. 

"  Sell.  1.   Application. 
Sch.  2.  Applied  to  triangles. 
Sell.  3.  Angle  of  equiangular  poly- 

Prop.  XVI.  Sum  of  exterior  angles. 

f  Prop.    XVII.  Equilateral  inscribed,  regular. 
Prop.  XVIII.  Equiangular  inscribed  j  Sch.  Fails  for 

if  odd  No.  of  sides.      (  even  No. 
Prop.      XIX.  Equiangular  circumscribed,  regular. 
Prop.       XX.  Equilateral    circbd.    if  J  Sch.  Fails  for 
odd  No.  of  sides.        \  even  No. 


Def. 


Regular. 


Prop.     XXI 


L  Prop.  XXII. 
Convex  Broken  Line. 


r  Cor.  1.    Inscribed. 
Cor.  2.   Centres. 
Circf  can  be  cir-      Def.  Angle  at  cntr. 
cumscribed.        1  Cor.  3.  Value  of  an- 
gle at  centre, 
t  Bef.  Apothem. 
Side  of  inscribed  hexagon. 


Prop.  XXIII.  Convex  broken  line  < 
than  — . 


Exercises. 


Cor.  1.  Sum  of  two  sides  of  tri- 
angle. 

Cor.  2.  Diflf.  of  two  sides  of  ti'i- 
angle. 

Cor.  3.  Lines  from  point  within 
triangle. 


OF  EQUALITY.  '  121 


SECTION  VIII. 

OF  EQUALITY. 


277*  Equality  signifies  likeness  in  every  resjoect. 

278,  The  equality  of  magnitudes  is  usually  shown  by  applying 
one  to  the  other,  and  observing  that  they  coincide. 


PROPOSITION  I. 

279.  Tlieorem. — Two  straight  lines   of  the  same  length  are 
equal  magnitudes.* 

Dem.— Let  AB  and  CD  be  two  straight  lines  of  the  same  length ;  then  are 
they  equal. 

For,  conceive  the  extremity  C  of  CD  placed  at  A, 

and  the  other  extremity  somewhere  in  AB,  or  in  AB         /\ g 

produced,  as  tlie  case   may  be.    Now,  the  point 

which  traces  AB  passes  through  all  points  in  the         ^ D 

direction  of  B  from  A ;  and  hence,  if  CD  is  traced  Fig.  198.* 

from  A  towards  B,  it  will  pass  through  the  same 

points  as  far  as  they  mutually  extend.  The  lines  therefore  coincide,  as  far  as 
they  both  extend ;  and,  being  of  the  same  length,  D  falls  at  B,  and  they  coincide 
throughout ;  they  are,  therefore,  equal,    q.  e.  d. 

III. — The  truth  of  this  theorem  is  so  evident,  that „ 

the  student  may  fail  to  see  the  point  of  the  demonstra- 
tion.   Let  him  see  if  he  can  say  the  same  things  of        / 
two*curved  lines  A?/iB,  and  C/iD,  which  are  of  the     ^  rr  ^B 

same  length.  /"''"^^^X. 

The  substance  of  the  demonstration  is  as  follows :      _^/  \ 

A  line  has  two  properties,  and  only  two,  form  and 
magnitude.    Straight  lines,  being  of  the  same  form,  if  ^'°"  ^^^* 

they  are  of  the  same  magnitude,  are  alike  in  all  respects;  i.  e.,  they  are  equal. 
Now,  a  line,  as  a  magnitude,  has  only  one  dimension,  viz.,  length.  If^  there- 
fore, two  lines  have  the  same  length,  they  have  the  same  magnitude. 

•  *  See  Preface. 


122 


ELEMENTAEY  PLANE  GEOMETRY. 


PROPOSITION  II. 

2S0»  Theorem. — Two  circles  luJiose  radii  are  of  the  same  length 
are  equal ;  i.  e.,  the  circumferences  are  equals  and 
the  circles  equal 

Dem. — Let  there   be   two   circles  whose  radii  AB  and 
CD  are  of  the  same  length ;  then  are  tlie  circles  equal. 

For,  place  the  second  circle  on  the  first,  with  the  centre 
C  at  A,  and  CD  in  AB.  As  CD  =  AB,  D  will  fall  at  B. 
Now,  ever^'  point  in  the  plane  at  a  distance  AB  trom  A  is  in 
the  circumference  of  circle  A.  But  every  point  at  a  distance 
CD  from  the  common  centime  is  in  the  circumference  of 
circle  C.  Hence,  the  two  figures  coincide,  and  the  circles 
are  alike  in  all  respects,  i.  e.^  are  equal,  q.  e.  d. 
Fig.  200. 


OF   ANGLES. 


PROPOSITION  ni. 

281.  Theorem, — Two  angles  whose  sides  are  par  allele  two  and 
two,  and  lie  in  the  same  or  in  opposite  directions  from  their  vertices, 
are  equal. 

Dem.— 1st  In  (a)  or  (a)  let  B  and  E  have  BA  and  ED  parallel,  and  extending 

in  the  same  direction  from  the 
vertices,  and  also  BC  and  EF; 
then  are  B  and  E  equal.  For, 
produce  (if  necessary)  either  two 
sides  which  are  not  parallel,  till 
they  intersect,  as  at  H ;  then  are 
the  corresponding  angles  DHC  and 
DEF,  and  DHC  and  ABC  equaJ 
{152).    Hence,  ABC  =  DEF. 

2nd.  In  (6)  and  (6)  let  B'  and  E' 
have  B'A'  parallel  with  E'F',  but 
extending  in  an  opposite  direction 
from  its  vertex ;  and  in  like  manner 
B'C  parallel  with,  but  extending  in 
For,  produce  (if  neces- 


C 
Fig.  201, 
an  opposite  direction  from  PD';  then  are  B'  and  E'  equal 


sary)  two  of  the  sides  which  are  not  parallel  till  they  intersect,  as  at  H';  then 
D'H'B'  =  the  corresponding  angle  D'E'F',  and  also  =  the  alternate  interior 
angle  A'B'C ;  whence  A'B'C'  =  D'E'F'.    q.  E.  D. 


EQUALITY  OF  ANGLES. 


123 


PROPOSITION  IV. 

282*  Theorem, — Ifhvo  angles  have  hoo  sides  parallel  and  ex- 
tending m  the  same  direction  vnth  each  other,  while  the  other  two 
sides  are  parallel  and  extend  in  opposite  directions  from  each  other, 
the  angles  are  sup>plemental. 


Dem.— Let  ABC  and  DEF  be  two  angles, 
having  BC  and  ED  parallel,  and  extending 
in  the  same  direction  from  the  vertices, 
and  AB  and  EF  parallel,  and  extending  in 
opposite  directions  from  the  vertices ;  then  are 
ABC  and  DEF  supplements  of  each  other. 

For,  produce  the  two  sides  not  parallel,  if 
necessary,  till  they  meet.  Now,  BHD  is  the 
supplement  of  BHE  by  {131),  BHE  =  the  al- 
ternate interior  angle  DEF,  and  BHD  =  the 
corresponding  angle  ABC.  Therefore,  ABC  is 
the  supplement  of  DEF.     q.  e.  d. 

[This  demonstration  is  adapted  to  the  upper 
cut;  let  the  student  adapt  it  to  the  lower.] 


PROPOSITION  T. 

283,  Theorem, — Iftioo  angleshave 
their  sides  respectively  perpendicular  to 
each  other,  the  angles  are  either  equal  or 
supplementary. 

Dem, — Let  BA  be  perpendicular  to  EF  or 
to  E'F',  and  BC  to  ED ;  then  is  ABC  =  DEF. 
For,  through  B  draw  BO  and  BN,  respectively 
parallel  to  ED  and  EF;  then  by  the  preceding 
jTropositions  NBO  =  DEF,  and  is  the  supple- 
ment of  F'E'D.  But  NBA  =:  OBC,  since  both 
are  right  angles.  Take  away  OBA  from  each, 
and  we  have  NBO  =  ABC;  and  as  NBO  is  the 
supplement  of  F'E'D,  ABC  is  also  the  supple- 
ment of  F'E'D.     Q.  e.  d. 


124 


ELEltENTABY  PLANE  GEOMETEY. 


OF  TRIMGLES. 


PROPOSITION  VI. 

284.  Theorem. — Tiuo  tricmgles  ivMcli  have  hoo  sides  and  the 
included  angle  of  one  equal  to  two  sides  and  the  included  angle  of  the 
other,  each  to  each,  are  equal, 

Dem.— Let  ABC  and  DEF 

be  two  triangles,  having 
AC  =  DF,  AB  =  DE,  and 
angle  A  =  angle  D ;  then 
are  the  triangles  equal. 

For,  place    the    triangle 
ABC  in  the  position  (6),  the 
side  AB  in  its  equal  DE,  and 
the  angle  A  adjacent  to  its 
equal  angle  D.     Then   re- 
volving ABC  upon  DB,  until 
it  falls  in  the  plane  on  the 
opposite  side  of  DB,  since  angle  A  =  angle  D,AC  will  take  the  direction  DF  ; 
and  as  AC  =  DF,  C  will  fall  at  F.     Hence  BC  will  fall  in  EF,  and  the  triangles 
will  coincide.     Therefore  the  two  triangles  are  equal,    q.  e.  d. 

We  may  also  make  the  application  of  ABC  to  DEF  directly,  as  in  {85).  The 
method  here  given  is  used  for  the  purpose  of  uniformity  in  this  and  the  follow- 
ing. We  may  observe  that  in  this,  as  in  the  other  cases,  DB  is  perpendicular  to 
PC,  and  bisects  it  at  0.  This  fact  might  easily  be  shown,  and  the  demonstra- 
tion be  based  upon  it 

2S5.  ScH. — This  proposition  signifies  that  the  two  triangles  are  equal  in  all 
iespects,  1.  e.,  that  the  two  remaining  sides  are  equal,  as  CB  =  FE;  that  angle 
C  =  angle  F,  angle  B  =  angle  E,  and  that  the  areas  ai'e  equal 


Fig.  204. 


PROPOSITION  TIL 

286.  Theorem. — Tiuo  triangles  ivhich  have  two  angles  and  the 
included  side  of  the  one  equal  to  two  angles  a7id  the  included  side  of 
the  other,  each  to  each,  are  equal. 


EQUALITY  OF  TRIANGLES. 


125 


Pig.  205. 


Dem.— Let  ABC  and  DEF 

be  two  tiiuDgles,  having 
angle  A  =  angle  D,  angle 
B  =  angle  E,  and  side  AB 
=  side  DE;  then  are  the 
triangles  equal. 

For,  place  ABC  in  the 
position  (b),  the  side  AB  in 
its  equal  DE,  the  angle  A 
adjacent  to  its  equal  angle  D> 
and  B  adjacent  to  its  equal 
angle  E.  Then  revolving 
ABC  upon  DB  till  it  falls  in  the  plane  on  the  same  side  as  DFE,  since  angle  A  = 
angle  D,  AC  will  take  the  direction  DF,  and  C  will  fall  somewhere  in  DF  or 
DF  produced.  Also,  since  angle  B  =  angle  E,  BC  will  take  the  direction  EF, 
and  C  will  fall  somewhere  in  EF,  or  EF  produced.  Hence,  as  C  falls  at  the 
same  time  in  DF  and  EF,  it  falls  at  their  intersection  F.  Therefore  the  two 
triangles  coincide,  and  are  consequently  equal,    q.  e.  d. 

287,  Cor. — //  one  triangle  has  a  side,  its  opposite  angle,  and  one 
adjacent  angle,  equal  to  the  corresponding  parts  in  another  triangle, 
each  to  each,  the  triangles  are  equal. 

For  the  third  angle  in  each  is  the  supplement  of  the  sum  of  the  given  angles, 
and  they  are  consequently  equal.  Whence  the  case  is  included  in  the  pro- 
position. 

288*  ScH. — A  triangle  may  have  a  side  and  one  adjacent  angle  equal  to  a 
side  and  an  adja- 
cent angle  in 
another,  and  the 
second  adjacent 
angle  of  the  first 
equal  to  the  angle 
opposite  the  equal 

side  in  the  second,  and  the  triangles  not  be  equal.  Thus,  in  the  figure,  AB  := 
C'^',  A  =  A',  and  B  =  B' ;  but  the  triangles  are  evidently  not  equal.  [Such 
triangles  are,  however,  similar,  as  will  be  shown  hereafter.] 


® 


PROPOSITION  rm. 


289.  Theore^n, — Two  triangles  ivhich  have  two  sides  and  an 
angle  opposite  one  of  these  sides,  in  the  one,  equal  to  the  corresponding 


126 


ELEMENTARY  PLANE  GEOMETRY. 


parts  ill  the  other,  are  equal,  if  of  these  tiuo  sides  the  one  opposite  the 
giveii  a?igle  is  equal  to  or  greater  than  the  one  adjacent. 

Dem.— In  the  triangles  ABC  and  DEF,  let  AC  =  DF,  CB  =  FE,  A  =  D,  and 

CB  {=  FE)  =  AC  {=  DF);  then  are  the  triangles  equal. 

For,  apply  AC  to  its  equal 
C  F  DF,  the  point  A  falling  at 

D  and  C  at  F.  Since  A  = 
D,  AB  will  take  the  direc- 
tion DE.  Let  fall  the  per- 
pendicular FH  upon  DE,  or 
DE  produced.  Now,  CB 
being  ^  DF,  cannot  fall 
between  it  and  the  perpen- 
dicular, but  must  fall  in  FD 
or  beyond  both.  As  there 
can  be  but  one  line  on  the 
same  side  of  the  perpen- 
dicular equal  to  CB,  and  as 

FE  =  CB,  CB  must  fall  in  FE.    Hence,  the  two  triangles  coincide,  and  are 

consequently  equal.     Q.  e.  d. 


290.  ScH.  l.-If  A  and  D  are  acute  and  CB  {=  FE)  =  AC  {=  DF),  the  tri- 
angles are  isosceles.  If  A  and  D  are  right  or  obtuse,  CB  (=FE)  must  be  greater 
than  AC  (=  DF),  in  order  that  there  may  be  a  triangle,  since  the  right  or  obtuse 
angle  is  the  greatest  angle  in  a  triangle,  and  the  greatest  side  is  opposite  the 
greatest  angle.    This  impossibility  appears  also  from  the  demonstration  above. 


291.  ScH.  2.— If  A  and  D  are  a€ui£,  and  the  side  opposite  A,  i.  e.,  CB,  is  less 

than  AC,  it  must  be   equal   to  or 
greater  than  the  pei-pendicular  CI 
(=  FH)  in  order  to  have  a  triangle. 
Then,  applying  AC  to  DF,  and  ob- 
serving that  AB  takes  the  direction 
DE,  and  that  EF,  which  =  CB,  being 
intermediate  in  length  between  DF 
and  FH,  may  lie  on  either  side  of 
FH,  we  see  that  ABC  may  or  may 
not  coincide  with  DEF.     "Whether  it  does  or  not  will  depend  upon  whether 
angle  C  =  angle  F,  or  whether  AB  =  DE.     This  is  the  ajibiguous  case  in 
the  solution  of  triangles,  and  should  receive  special  attention. 


EQUALITY  OF  TRIANGLES. 


127 


PROPOSITION  IX. 

292,17ieore^n. — Tiuo  triangles  wliich  have  the  three  sides  of 
the  one  equal  to  the  three  sides  of  the  other,  each  to  each,  are  equal. 


Dem.— Let  ABC  and  DEF  be  two  triangles,  in  whicli  AB  =  DE,  AC  =  DF, 
and  BC  =  EF;  then  are  tlie 
triangles  equal. 

For,  place  the  triangle 
ABC  in  the  position  (&)^  and 
the  side  AB  in  its  equal  DE, 
so  that  the  other  equal  sides 
shall  be  adjacent,  as  AC  ad- 
jacent to  DF,  and  BC  to  EF. 
Draw  FC.  Now,  since  DC 
=  DF,  and  EC  =  EF,  DB  is 
perpendicular  to  FC  at  its 
middle  point  {130).  Hence,  p^^  209 

revolving  ABC  upon  DB,  it 

will  coincide  with  DEF  when  brought  into  the  plane  of  the  latter     Therefore 
the  two  triangles  are  equal,     q.  e.  d. 

293,  Cor. — In  tioo  equal  triangles,  the  equal  angles  lie  opposite 
the  equal  sides. 

294,  Sen. — If  the  triangles  compared,  as  in  the 
three  preceding  propositions,  have  an  obtuse  angle,  and 
the  two  sides  first  brought  together  are  sides  about  the 
obtuse  angle,  the  figure  will  take  the  form  in  the  mar- 
gin ;  but  the  demonstration  will  be  the  same.  When 
the  three  sides  are  the  given  equal  parts,  the  form  of 
figure  given  in  the  demonstration  above  can  always  be 
secured  by  bringing  together  the  two  greatest  sides. 

Fio.  210. 


PROPOSITION  X. 

29S,  Theorem, — If  tioo  triangles  have  tiuo  sides  of  the  one 
respectively  equal  to  tiuo  sides  of  the  other,  and  the  included  angles 
unequal,  the  third  sides  are  unequal,  and  the  greater  third  side 
belongs  to  the  triangle  having  the  greater  i^icluded  angle. 


VzS 


ELEMENTARY  PLANE   GEOMETRY. 


Dem.— Let  ACB  and  DEF  be  two  tri- 
angles having  AC  =  DF,  CB  =  FE,  and 
C  >  F;  then  is  AB  >  DE. 

For,  placing  the  side  DF  in  its  equal 
AC,  since  angle  F  <  angle  C,  FE  will  fall 
within  the  angle  ACB,  as  in  CE.  Then  let 
the  triangle  ACE  =  the  triangle  DFE.  Bi- 
sect ECB  with  CH,  and  draw  HE.  The 
triangles  HCB  and  HCE  have  two  sides 
and  the  included  angle  of  the  one,  respec- 
tively equal  to  the  corresponding  parts  of 
the  other,  whence  HE  =  HB.  Now  AH  -i- 
HE  >  AE;  but  AH  +  HE  =  AH  +  HB  =r 
AB.     Therefore,  AB  >  AE.     q.  e.  d. 

296.   Cor.  —  Conversely,  If  two 

-P^  211  ^^^^^  ^f  ^'^^  triangle  are  respectively 

equal  to  tivo  sides  of  another^  and  the 

third  sides  iinequah  the  aJigle  opposite  this  third  side  is  the  greater 

in  the  triangle  ivhich  has  the  greater  third  side. 

Dem.— If  AC  =  DF,  CB  =  FE,  and  AB  >  DE,  angle  C  >  angle  F.  For,  if 
C  =  F,  the  triangles  would  be  equal,  and  AB  =  DE  {284);  and,  if  C  were  less 
than  F,  AB  would  be  less  than  DE,  by  the  proposition.  But  both  these  conclu- 
sions are  contrary  to  the  hypothesis.  Hence,  as  C  cannot  be  equal  to  F,  nor 
less  than  F,  it  must  be  greater. 


PROPOSmON  XL 

297.  Hieovein, — Tioo  right  angled  triangles  tohich  have  the 
hypotemise  and  one  side  of  tlw  one  equal  to  the  hypotenuse  and  one 
side  of  the  other,  each  to  eacJi,  are  equal. 

Dem.— In  the  two  triangles  ABC  and  DEF,  right  angled  at  B  and  E,  let  AC 
=  DF,  and  BC  =  EF;  then  are  the  triangles  equal. 

For,  place  BC  in  its  equal 
Q  PC  ^^>  so  that  tlie  right  angles 

shall  be  adjacent,  the  angles 
A  and  D  lying  on  opposite 
sides  of  EF.  as  in  {h\  Since 
E  and  B  are  right  angles, 
DA  is  a  straight  line.  Now, 
since  equal  oblique  lines,  as 
Fig.  212.  ^^  ^^fl  CA,    cut  off  equal 

distances  from  the  foot   of 
the  perpendicular  {14:1\  DE  =  BA;  and  revolving  CAB  upon  FB,  the  two 


EQUALITY  OF  QUADKILATERALS.  129 

triangles  will  coincide  when  CAB  falls  in  the  plane  on  the  side  D.    Therefore, 
the  triangles  are  equal,     q.  e.  d. 


PROPOSITION  xn. 

298.  TJieorem, — Two  right  angled  triangles  liaving  the  hypo- 
tenuse and  one  acute  angle  of  the  one  equal  to  the  hypotenuse  and 
an  acute  angle  of  the  other,  are  equal. 

Dem.— One  acute  angle  in  each  being  equal,  the  other  acute  angles  are 
equal,  since  they  are  complements  of  the  same  angle  {222).  The  case  is,  then, 
that  of  two  angles  (the  acute  angles  in  each),  and  their  included  side  (the  hy- 
potenuse), and  falls  under  {286). 


PROPOSITION  xni. 

299,  Tlieorem, — Two  right  angled  triangles  having  a  side  and 
a  corresponding  acute  angle  in  each  equal,  are  equal. 
This  also  falls  under  {286).    Let  the  student  show  why. 


OF  QUADRILATERALS. 


PROPOSITION  xiy. 

300,  Tlieorem, — T2V0  quadrilaterals  are  equal  when  the  follow- 
ing parts  are  equal,  each  to  each,  in  both  quadrilaterals,  and  similarly 


1.  The  triangles  into  which  either  diagonal  divides  the  quadrilaterals. 

2.  The  four  sides  and  either  diagonal. 

3.  The  four  sides  and  one  angle. 

4.  Three  sides  and  the  two  included  angles. 

5^.  Three  angles  and  any  tioo  sides,  if  the  other  two  sides  are  non- 
parallel. 

Dem. — 1.  This  case  is  demonstrated  by  applying  one  quadrilateral  to  the  other. 

2.  This  case  is  reduced  to  the  former  by  {292). 

3.  Drawing  the  diagonal  opposite  the  known  angle,  this  case  is  reduced  to 
the  first  by  {284),  and  {2i)2). 

4.  This  is  demonstrated  by  applying  one  quadrilateral  to  the  other  ;  or,  draw- 
ing a  diagonal,  it  may  be  reduced  to  case  first  by  {284). 

5.  In  this  case  the  quadrilaterals  are  mutually  equiangular  by  {25o).    If, 
then,  the  two  Bides  are  adjacent,  by  drawing  the  diagonal  joining  their  extremi- 

9 


130  ELEMENTARY  PLAN"E  GEOMETRY. 

ties  the  case  is  brought  under  the  first  by  (284),  and  (280).     If,  however,  the 
y\  ^  two  known  sides  are  non-adjacent,  by  pro- 

^y    \  /  \         ducing  the  unknown  sides  two  triangles 

are  formed  in  each  figure,  which  are  mu- 
tually equal  by  (280),  and  the  case  comes 
under  the  axiom  concerning  equals  sub- 
^    2lT  '     '    ^     tracted  from  equals. 

[Let  the  pupil  draw  the  figures  and  give 
the  demonstrations  in  full  form.  Trapeziums  should  be  used ;  although  it 
should  be  seen  in  each  case — except  the  fifth— that  the  truth  applies  to  any 
other  form  of  quadrilateral.] 


PROPOSITION  XT. 

301,  Theorem. — Two  2)arallelograms  having  tivo  sides  and  the 
incUided  angle  of  the  one  equal  to  tioo  sides  and  the  included  angle 
of  the  other,  each  to  each,  are  equal. 

Dem.— Let  AC  and  EG  be  two  parallelograms,  with  AD  =:  EH,  AB  =  EF,  and 

A  =  E ;  then  are  they  equal. 

For,  applying  the  angle  E  to  A,  since  EH 
^  AD,  H  will  fall  at  D ;  and  since  EF  =  AB, 
F  will  fall  at  B.  Now,  through  D  but  one 
Hue  can  be  drawn  parallel  to  AB ;  hence  HG 
will  fall  in  DC,  and  C  will  be  found  in  DC, 
or  in  DC  produced.  In  like  manner,  since  but 
one  parallel  to  AD  can  be  drawn  through  B, 
YiQ^  214.  ^^  must  fall  in  BC,  and  G  be  found  in  BC, 

or  in  BC  produced.     Therefore,  as  G  falls  at 
the  same  time  in  DC  and  BC,  it  falls  at  C,  and  the  parallelograms  coincide. 

302,  Cor. — Two  rectangles  of  the  same  base  and  altitude  are 
equal. 


/ 

7 

/ 

7' 

A 

B 

hi 

r' 

/ 

/ 

/ 

7G 

/ 

/ 

/ 

L 

F 

OF  POLYGONS. 


PROPOSITION  XVI. 

S03.  Theorem. — Two  polygons  of  the  same  number  of  sides, 
having  all  the  parts  of  the  one  except  three  angles,  known  to  he  respec- 
tively equal  to  the  corresponding  parts  of  the  other ^  are  equal. 

Dem. — If  the  two  polygons  AE*  and  A'E',  have  all  the  parts  of  the  one  equal 
to  the  corresponding  parts  of  the  other,  each  to  each,  except  three  angles ;  then 
are  the  pol5''gons  equal. 

*  It  is  often  more  convenient  to  read  a  polygon  by  two  letters,  instead  of  all  those  at  the 
vertices. 


EQUALITY   OF   POLYGO>'S. 


131 


Fig.  215. 


1st.  When  the  three  un- 
known angles  are  consecu- 
tive, as  G,  F,  E,  and  C,  F', 
E'.  Draw  CE,  and  G'E'. 
Apply  polygon  A'E'  to  AE, 
beginning  with  g'  in  its 
equal  ^r;  A'  =  A,  and  a'  = 
a;  hence,  B'  falls  at  B:  B' 
=  B,  and  b'  =  b,  hence,  C 
falls  at  C ;  etc.  Thus,  we 
may  show  that  the  perime- 
ters coincide  till  we  reach  E'  and  E.  Then  will  G'E'  =  GE,  and  the  triangles 
GFE  and  G'F'E',  having  their  corresponding  sides  respectively  equal,  are  them- 
selves equal,  and  the  polygons  coincide  throughout. 

2d.  When  two  of  the  unknown  angles  are  consecutive,  and  the  third  not  con- 
secutive with  these,  as  G,  E,  D,  and  G',  E',  D'.  From  the  angle  which  is  not 
consecutive  with  the  other  two,  draw  diagonals  to  the  other  angles,  as  GE,  GD, 
and  G'E',  G'D'.  Now,  G'A'B'C'D'  can  be  applied  to  GABCD,  and  G'F'E'  to  GFE, 
in  the  ordinary  way.  Hence,  the  triangles  G'E'D'  and  GED  are  mutually  equi- 
lateral, and  consequently  equal.    Therefore  the  polygons  are  equal. 

3d.  When  no  two  of  the  three  unknown  angles  are  consecutive,  as  G,  B,  D, 
and  G',  B',  D'.  Join  the  unknown  angles  by  diagonals,  as  GB,  GD,  BD,  and  G'B', 
G'D',  B'D'.  Now,  polygon  G'F'E'D'  can  be  applied  to  GFED,  D'C'B'  to  DCB, 
and  G'A'B'  to  GAB  in  the  ordinary  way.  Hence,  the  triangles  G'D'B'  and  GDB 
are  mutually  equilateral,  and  consequently  equal.  Therefore  the  polygons  are 
equal.* 

S04,  Cor. — Two  quadrUaterals  having  their  corresponding  sides 
eqitaly  and  an  angle  in  one  equal  to  the  corresponding  angle  in  the 
other,  are  eqtial. 


PROPOSITION  xyn. 

SOo,  Theorem, — Tivo  polygons  of  the  same  number  of  sides, 
having  all  the  parts  of  the  one  except  tico  angles  and  the  i^icluded 
si^e,  hnown  to  he  respectively  equal  to  the  corresponding  parts  of  the 
other,  are  equal. 

Dem.— Let  the  unknown  parts  be  C,  c,  D,  and  C,  c' ,  D'.  From  any  other  two 
of  the  mutually  equal  angles,  as  G  and  G',  draw  the  diagonals  GO,  GD,  GX',  G'D', 


*  Notice  that  in  each  case  the  unknown  angles  are  to  form  the  vertices  of  triangles,  which 
the  argument  shows  to  be  equilateral,  and  therefore  equal.  In  Case  1st,  we  have  to  draw  only 
one  line  in  order  to  give  the  triangles,  as  two  sides  are  sides  of  the  polygon ;  in  Case  2d,  we 
have  to  draw  two  sides ,  and  in  Case  3d,  three  sides,  for  analogous  reasons. 


133 


ELEMENTARY  PLANE  GEOMETET. 


to  the  unknown  angles.  Then 
the  polygon  G'F'E'D'  can  be 
applied  in  the  ordinary  way 
to  GFED,  /'  being  placed  in 
/,  etc     So  also  G'A'B'C  can 
be  applied  to   GABC,  begin- 
ning with  g'  in  its  equal  g. 
Hence,   angle    F'G'D'  =  FGD, 
A'G'C'=AGC;    and,   adding, 
F'G'D'  +  A'G'C  =  FGD  + 
AGO.      Subtracting      these 
equals  from  G'  =  G,  we  have  C'G'D'  =  CGD.    Whence  the  triangles  C'G'D'  and 
CGD  have  two  sides  and  their  included  angle  equal  in  each,  and  are  equal; 
therefore  the  polygons  are  equal  in  all  their  parts. 


SOG,  ScH.— When  the  unknown 
angles  are  both  separated  from  the 
unknown  side,  the  polygons  may  or 
may  not  be  equal — the  case  is  am- 
biguous. Thus,  if  C  and  E  are  the 
unknown  angles  and  AH  the  un- 
known side,  the  polygons  ABCDEFG, 
and  A'B'C'D'EFG  fulfill  the  condi- 
tions, but  are  not  equal.  By  draw- 
ing CE,  CA,  and  EH,  the  case  is  re- 
|-1  duced  to  that  of  two  quadrilaterals 
having  all  the  parts  equal,  each  to 
each,  except  two  angles  and  their 

non  adjacent  side  ;  in  which  case  the  quadrilaterals  are  not  necessarily  equal. 
So,   also,   when  one  of  the  unknown  angles  is  adjacent  to  the  unknown 

side  and  the  other  separated,  the  polygons  may  or  may  not  be  equal.    Thus,  let 


the  unknown  parts  be  D,  c,  C,  and  D',  c' ,  C.  From  the  separated  angle  draw  the 
diagonals  to  the  extremities  of  the  unknown  side,  as  GC,  CD  (or  GDi),  and  C'C, 
CD'.     In  the  usual  way  G'A'B'C'  can  be  applied  to  GABC,  and  G'F'E'D'  to 


EQUALITY  OF  POLYGONS. 


133 


CFED.  Whence  C'C  =  GC,  CD'  =  CD,  and  angle  C'C'D'  =  CCD.  Thus  the 
case  is  reduced  to  that  of  two  triangles  having  two  sides  and  an  angle  oppo- 
site one  of  them  mutually  equal,  and  is,  therefore,  ambiguous.  The  polygon 
(a)  may  have  the  part  corresponding  to  C'F'E'D'  situated  as  CFED,  or  as 
CFiEiDi,    In  the  former  case  the  polygons  are  equal,  in  the  latter  not. 

307*  Cor. — Two  quadrilaterals  having  three  sides  and  the  corre- 
sponding angles  included  by  these  sides  equal,  are  equal. 

This  falls  under  the  1st  case. 

308*  ScH. — If  the  three  unknown  or  excepted  parts  are  all  sides,  the  poly- 
gons are  not  necessarily  equal,  as  will  appear  by  an  inspection  of  the  figure.    The 


Fig.  219. 


unmarked  sides  being  the  excepted  ones,  the  polygons  may  be  those  included  by 
the  continuous  lines,  or  those  included  in  part  by  the  broken  lines,  all  the  parts 
being  equal  in  each  two,  except  the  three  unknown  ones. 


PROPOSITION  xvni. 

309,  Theorem, — Two  iMijgons  of  the  same  numler  of  sides, 
having  tiuo  adjacent  sides  and  the  diagonals  drawn  from  the  included 
angle,  in  the  one,  respectively  equal  io  the  corresponding  parts  in  the 
other,  and  their  corresponding  included  angles  equal,  are  equal 
figures. 

Dem.— The  demonstration  is  based  upon  {284).  Let  the  student  draw  the 
figures,  and  make  the  applications. 


PROPOSITION  XIX. 
310.  Theorem. — Two  polygons  of  the  same  number  of  sides, 
having  all  the  parts  {sides  and  angles)  of  the  one  respectively  equal 
to  the  corresponding  parts  of  the  other,  exce2^t  two  parts,  are  equal,  un- 
less the  exapled  partA  arc  parallel  sides. 


134 


ELEMENTARY  PLANE  GEOMETRY. 


Dem. — The  demonstration  can  be  supplied  by  the  pnpil,  as  it  i3  similar  to  the 
several  preceding.  The  cases  will  be,  1st,  When  two  angles  are  excepted, 
(a)  they  being  consecutive,  {b)  they  not  being  consecutive  ; — 2d,  An  angle  and  a 
side,  (a)  consecutive,  {b)  not  consecutive  ; — 3d,  Two  sides,  (a)  consecutive,  {b)  not 
consecutive. 


EXERCISES. 

1.  JProb. — Having  two  sides  and  their  included  angle  given,  to 
construct  a  triaiigle. 

Sug's. — The  student  should  draw  two  lines  on  the  blackboard,  and  a  detached 
angle,  as  the  given  parts.  Then,  making  an  angle  equal  to  the  given  angle 
{200),  he  should  lay  off  the  given  sides  from  the  vertex  on  the  sides  of  the 
angle,  and  join  their  extremities.  The  triangle  thus  formed  is  the  one  required, 
for  any  other  triangle  formed  with  these  two  sides  and  this  angle  will  be  just 
like  this  by  {284). 

2.  I^i*oh, — Having  two  angles  and  their  included  side  give?i,  to 
C07istruct  a  triangle. 

3.  JProb, — Having  the  three  sides  of  a  triangle  given,  to  construct 

the  triayigle. 

Solution. — Let  a,  b,  and  c,  be  the  given 
sides.  Draw  an  indefinite  line  CX,  and  on 
it  take  CB  —  a.  From  C  as  a  centre  with 
5  as  a  radius,  describe  an  arc  as  near  as  can 
be  discerned  where  the  angle  A  will  fall. 
From  B,  with  a  radius  c,  describe  an  arc 
intersecting  the  former.  Then  is  ABC  the 
triangle  required,  since  any  other  triangle 
having  the  same  sides  would  be  equal  to 
ABC  {292). 

4.  ^roh. — To  inscrihe  a  circle  in  a  given  triangle. 

Solution. — For  the  method  of  doing  it  see  Part  I.  {70).    To  prove  the 

method  correct,  we  observe  that  the  triangles 
ODB  and  QBE  have  OB  common,  and  are 
mutually  equiangular  ;  hence  they  are  equal, 
and  CD  =  OE.  In  like  manner  triangle 
OEC  =  OFC,  and  OE  =  OF.  [Tiiangle  OFA 
=  ODA ;  but  we  do  not  need  the  fact  in  the 
demonstration.]  Since  OD  =  OE  =  OF,  the 
circumference  stnick  from  O  as  a  centre  with 
a  radius  OD,  passes  through  E  and  F.  More- 
over, since  each  side  of  the  U'iangle  is  per- 
pendicular to  a  radius  at  its  extremity,  it  is  tangent  to  the  circle  {172) ;  and 
the  circle  is  inscribed. 


Fig.  2iJ0. 


OP  EQUALITY. 


135 


5.  JProb, — Having  tivo  sides  and  an  angle  opposite  one  of  them 
given,  to  construct  the  triangle. 

Solution. — 1st.  When  the  given  angle  is  right  or  obtuse,  the  side  opposite 
must  be  greater  than  the  side  adjacent,  as  the  greatest  side  is  opposite  the 
greatest  angle  {224^),  and  the  greatest  angle  in  such  a  triangle  is  Xu.^  right  or 


obtuse  angle.  In  this  case  let  m  and  o  be  the  given  sides,  and  0  the  angle  oppo- 
site o.  Draw  an  indefinite  line  O'X,  construct  0'  equal  to  0,  and  take  O'N' 
equal  to  m.  .  From  N'  as  a  centre,  with  a  radius  equal  to  0,  describe  an  arc  cut- 
ting O'X,  as  at  M'.  Draw  N'M'.  Then  is  N'M'O'  the  triangle  required,  since 
all  triangles  having  their  corresponding  parts  equal  to  w',  o\  and  0'  are  equal. 

2d.  When  the  given  angle  is  acute,  as  A,  there  will  be  no  solution  if  the 
given  side,  a,  opposite  A,  is  less  than  the  perpendicular ;  one  solution  if  a  =  p, 
or  if  a  >  than  both  j)  and  5,  and  two  solutions  if  a  >  ^,  and  less  than  h.  This 
will  appear  from  the  construction,  which  is  the  same  as  in  Case  1st. 

6.  If  a  perpendicular  be  let  fall  from  the 
right  angle  C  of  the  triangle  ACB  upon  the 
hypotennse,  as  CD,  show  from  {222)  that 
the  three  triangles  in  the  figure  are  mutually 
equiangular. 

7.  Given  the  sides  of  a  triangle,  as  15,  8,  and  5,  to  construct  the 
triangle. 

8.  Given  two  sides  of  a  triangle  a  =  20,  Z»  =  8,  and  the  angle  B 
opposite  the  side  h  equal  |  of  a  right  angle,*  to  construct  the  triangle. 

9.  Same  as  in  the  8th,  except  h  =  12.     Same,  except  that  i  =  25. 

10.'  Construct  a  triangle  with  angle  A  =  f  of  a  right  angle,  angle 
B  =  4-  of  a  right  angle,  and  side  a  opposite  angle  A,  15. 

11.  Construct  a  right  angled  triangle  whose  hypotenuse  is  16,  and 


♦  To  construct  this  angle,  bisect  an  angle  of  an  equilateral  triangle. 


136 


ELEMENTARY   PLANE   GEOMETRY. 


one  of  the  otlier  sides  7.  The  same  with  one  acute  angle  f  of  a 
right  angle,  and  a  side  about  the  right  angle  12.  Will  there  be  any 
difference  in  the  shape  of  the  triangles  if  one  is  constructed  with  the 
given  angle  adjacent  to  the  given  side,  and  the  other  with  it  oppo- 
site ?     Will  there  be  any  difference  in  the  size  9 

12.  Construct  a  right  angled  triangle  having  its  hypotenuse  20, 
and  one  acute  angle  ^  of  a  right  angle. 

13.  Construct  a  quadrilateral  three  of  whose  sides  are  20,  12,  and 
15,  and  the  angle  included  between  20  and  the  unknown  side  |  of  a 
right  angle,  and  that  between  15  and  the  unknown  side  -}  a  right 
angle. 

f  of  a  right  angle,  and  h  =  20.  From  D  as  a  centre,  with 
a  radius  12,  strike  the  arc  on.  At  any 
point  on  side  a,  make  an  angle  B'  = 
^  a  right  angle.  Take  B'7n  =  15,  and 
draw  Cm  parallel  to  AB'.  Fi*om  the 
intersection  C  draw  CB  parallel  to 
mS'.  Draw  CD.  Then  is  ABCD  the 
quadrilateral  required. 

Queries. — If  d  +  cis  less  than  the 
perpendicular  from  D  upon  AB,  then 
what  ?  If  equal  to  the  perpendicular, 
then  what?  Is  it  necessary  to  consider  angle  B  in  answeriug  the  two  pre- 
ceding queries  ? 

14.  Construct  a  parallelogram  whose  two  adjacent  sides  are'  6  and 
8,  and  whose  included  angle  equals  1^  right  angles. 

15.  Construct  a  heptagon  whose  sides  in  order  are  a  ==  4:,  b  =  6, 
c  =  5,  d  =  6,e  =  6,f  =  d,  (/  =  4:;  and  the  angle  included  between 
a  and  b,  1^  right  angles;  between  b  and  c.  If ;  c  and  d,  IJ;  d  and 

Sug's.— See  Fig.  187.  Proceed  in  ordei,  laying  off  the  parts  as  given,  from  A 
to  F.  Draw  AF.  From  F  as  a  centre,  with  a  radius/  =  3,  strike  an  arc,  and  also 
from  A,  with  a  radius  g  =  4.    The  intersection  of  these  arcs  will  determine  C. 

Queries.— ^\\a.i  is  tlie  limit  of  the  sum  of  the  possible  values  of  the  given 
angles  ?  "What  the  limit  of  the  sum  of  tne  sides  included  between  the  unknown 
anglwi  ? 


OF  EQUALITY. 


137 


SYNOPSIS. 


f  WUat?    How  shown? 
Prop.  I.  Of  straight  lines. 
Prop.  II.  Of  circles. 


it 


i 


Prop.  III.  Sides  parallel.    Direction  same  or  opposite. 

Prop.  IV.      "  *'  "        one  same,  other  opposite. 

Prop.  V.        "     perpendicular. 

Prop.  VI.  Two  sides  and  included  angle.  ■{  Sch.  All  parts  equal. 

Prop.  VIL  Two  angles  and  (  ^'='^-  ^^'l^'  ^^^  fcljacent  and  one  oppo- 

mcluded  side,  j  .  ;   -p  '      f^^""  '''^'''''^' 
{  Sell.  Exception. 

Prop.  VIII.  Two  sides  and  angle  j  Sc7i.  1.  When  isosceles. 

opposite  one.  (  Sch.  2.  Wlien  ambiguous. 

T>r,^T.  TV    Ti  ,.^^  .•  i^o  i  (^^^'-  Equal  angles  opposite  equal  sides. 
Prop.  IX.  Three  sides,  j  ^^;,  ^^^^  ^^  ^g^^^^^  i^j^g^^     ^.^^^^  ^^  ^.^ 

-Prop.  X.   Two   sides  equal,   included  angles  un- )  ^^^   Converse 

Prop.  XI.  Hypotenuse  and  one  side. 
Prop.  XII.  Hypotenuse  and  one  acute  angle. 
<*    I  Prop.  XIII.  Side  and  one  acute  angle. 

Prop.  XIV.  Three  sides  and  non-included  angles  equal. 

Prop.  XV.  Two  parallelograms  having  two  (  Co?:  Rectangles    of 


sides  and  the  included  angles 
equal. 


same      base 
and  altitude. 


Prop.  XVI.  Three  angles  excepted.  ■{  Cor.  Quadrilaterals. 

V1TTT    m  1  J  (  ScJi.  1.  The  ambiguous  case. 

!  Prop.  XVIL  Two  angles  and  one  \  ^^,,   Quadrilaterals. 
i  side  excepted.  ^  ^^^^  ^  ^liree  sides  excepted. 

Prop.  XVIII.  Two  sides  and  included  diagonals. 

Prop.  XIX.  Any  two  parts  excepted. 


Exercises. 


Prob.  In  a  triangle,  given  two  sides  and  included  angle. 
Prob.     "  "  "        angles  "  side. 

Prob.     "  "  "        sides  and  angle  opposite  one. 

Prob.     "  "  "  three  sides. 

Prob.  To  inscribe  a  circle  in  a  triaui^le. 


138  ELEMENTARY  PLANE  GEOMETRY. 


3 


SECTION  IX. 

OF  EQUIVALENCY  AND  AREA. 


Sll.  Equivalent  Fif/iires  are  such  as  are  equal  in  magni- 
tude. 


PROPOSITION  I. 

312.  TJieorenu — Parallelograms  having  equal  bases  and  equal 
aUitudes  are  equivalent. 

Dem. — Let  ABCD  and  EFCH  be  two  parallelograms  having  equal  bases,  BC 
and  FC,  and  equal  altitudes;  then  are  they  equivalent. 

A  E'  D  H'         E  H  *^^^'  P^^^^  FG  in  its  equal 

7     \ 7  J 7      BC ;  and,  since  the  altitudes 

\  /  /  /        are  equal,  the  upper  base  EH 

\      /  /  /         will  fall  in  AD  or  AO  pro- 

/ \/  / /  duced,  as   E'H'.     Now,   the 

^  C  F  G  two  triangles  AE'B  and  DH'C 

Fig.  223.*  are  equal,  because  the  three 

sides  of  the  one  are  respectively  equal  to  the  three  sides  of  the  other.  Thus  AB 
=  DC,  being  opposite  sides  of  the  same  parallelogram.  For  a  like  reason,  E'B 
=  H'C.  Also,  E'H'  =  BC  =  AD.  From  AH'  taking  E'H',  AE'  remains,  and 
taking  AD,  DH'  remains.  Therefore  AE'  =  DH'.  These  triangles  being  equal, 
the  quadrilateral  ABCH'  -  the  triangle  AE'B  =  ABCH'  -  DH'C.  But  ABCH' 
-  AE'B  =  E'BCH'  =  EFCH ;  and  AB^CH'  -  DH'C  =  ABCD.     Hence,  ABCD  = 

EFCH.      Q.  E.  D. 

313,  Cor. — Any  parallelogram  is  equivalent  to  a  rectangle  having 
the  same  base  and  altitude. 


PROPOSITION  n. 

314,  TJieorem, — A  triangle  is  equivalent  to  one-half  of  any 
parallelogram  having  an  equal  base  and  an  equal  altitude  with  the 
triangle. 


OF  EQUIVALENCY. 


139 


DKM.~Let  ABC  be  a  triangle.  Through  C  draw  CD  parallel  to  AB  ;  and 
through  A  draw  AD  parallel  to  BC.  Then  is 
ABCD  a  parallelogram,  of  which  ABC  is  one- 
half  {243).  Now,  as  any  other  parallelogram 
having  an  equal  base  and  altitude  with  ABCD 
is  equivalent  to  ABCD  {312),  ^BO  is  equiva- 
lent to  one-half  of  any  parallelogram  having 
an  equal  base  and  altitude  with  ABC.    q. 

E.  D. 

315,  Cor.  1. — A  triangle  is  equivalent  to  one-half  of  a  rcctamjle 
having  an  equal  base  and  an  equal  altitude  with  the  triaixgle. 

316.  Cor.  2. — Triangles  of  equal  lases  and  equal  altitudes  are 
equivalent,  for  they  are  halves  of  equivalent  parallelograms. 


Fig.  -ZiA. 


PROPOSITION  III. 

317,  Theorein, — The  square  described  on  a  line  is  equivalent  to 
four  times  the  square  described  on  half  the  line,  nine  times  the  square 
described  on  one-third  the  line,  sixteen  times  the  square  on  one-fourth 
the  line,  etc. 

Dem. — Let  AB  be  any  line.  Upon  it  describe  the  square  ABCD.  Bisect  AB, 
as  at«c?,  and  AD,  as  at  a.  Draw  dc  parallel  to  AD,  and  ah  parallel  to  AB.  Now, 
the  four  quadrilaterals  thus  formed 

are  parallelograms    by  construction,  ^  r    n 

hence  their  opposite  sides  and  angles  ' 

are  equal ;  and  as  A,  B,  C,  and  D  are 
right  angles,  and  ka  —  kd  =  dB  — 
bB  =  etc.,  the  four  figures  1,  2,  3,  4, 
are  equal  squares.  Hence  kdoa  =  i 
ABCD,  In  like  manner  it  can  be 
shown  that  the  nine  figures  into  which  ^^^-  ^^• 

the  square  on  A'B'  is  divided  by  draw- 
ing through  the  points  of  trisection  of  the  sides,  lines  parallel  to  the  other  sides, 
^re  equal  squares.     Hence  AV,  the  square  on  ^  of  A'B',  is  i  of  the  square 
A'B'C'D'.    The  same  process  of  reasoning  can  be  extended  at  pleasure,  show- 
ing that  the  square  on  i  a  line  is  -iV  the  square  of  the  whole,  etc. 


n 

o 

c 

4      \      3 

<7 

d\ 

/      i      2 

7i 

A 

d. 

ti 

a 


8  \  9 


G    .    5    \4r 


or      7n\ 

4    '.  2    \  3 


l/" 


y" 


PROPOSITION  IV. 

31S.  Tlieorem. — A  trapezoid  is  equivalent  to  t%vo  triangles 
having  for  their  bases  the  upper  and  lovjer  bases  of  the  trapezoid,  and 
for  their  common  altitude  the  altitude  of  the  traijezoid. 


140 


ELEMENTAKY  PLANE  GEOMETRY. 


Dem.— By  constructing  any  trapezoid,  and  drawing  cither  diagonal,   the 
student  can  show  the  truth  of  this  theorem. 


319.  Proh, 


PROPOSITION  Y. 

To  reduce  any  i}olygon  to  an  equivalent  triangle. 


Solution.— Let  ABCDEFbca  polygon  which  it  is  proposed  to  reduce  to  an 
equivalent  triangle.     Produce  any  side,  as  BC,  indefinitely.    Draw  the  diagonal 

EC   and    DH    parallel    to    it, 
^  Draw  EH.    Now,  consider  the 

triangle  CDE  as  cut  off  from 
the  polygon  and  replaced  by 
CHE.  The  magnitude  of  the 
polygon  will  not  be  changed, 
since  CDE  and  CHE  have  the 
same  base  CE,  and  the  same 
altitude,  as  their  vertices  lie  in 
DH  parallel  to  EC.  From  the 
polygon  thus  reduced  we  cut 
the  triangle  FHE,  and  replace 
it  by  its  equivalent  FHI,  by  drawing  the  diagonal  FH,  and  the  parallel  El.  In 
like  manner,  by  drawing  FB  and  the  parallel  AG,  we  can  replace  FBA  by  its 
equivalent  FGB.  Hence,  CFI  is  equivalent  to  ABCDEF.  It  is  evident  that  a 
similar  process  would  reduce  a  polygon  of  any  number  of  sides  to  an  equiva- 
lent triangle. 


AREA. 


PROPOSITION  TI. 

320,  Theorem, — The  area  of  a  rectangle  is  equal  to  the  product 
of  its  base  and  altitude. 

Dem. — Let  ABCD  be  a  rectangle,  then  is  its  area  equal  to  the  base  AB  multi- 
plied by  the  altitude  AC. 

If  the  sides  AB  and  AC  are  commensurable,  take 
some  unit  of  length,  as  E,  which  is  contained  a  whole 
number  of  times  in  each,  as  five  times  in  AC,  and 
eight  times  in  AB,  and  apply  it  to  the  lines,  dividing 
them  respectively  into  five  and  eight  equal  parts. 
A  f  2  s  ^  3  e  7  e^  Frcm  the  several  points  of  division  draw  lines  through 
Fig.  2-27.  ^^^^  rectangle  perpendicular  to   its  sides.     The  rect- 

angle will  be  divided    into    small    parallelogram^, 
which  are  all  equal  squares,  as  the  angles  arc  all  right  angles,  and  the-  sides  all 


O                E 

D 

&     i    :    :    i    i    i    ;    • 

c  — i  — i—l— J— h'.- .{/'—.— 
^  ^\      \      \      \      \      \     \ 

OF  AREA.  141 

equal  to  each  otlier.  Each  square  is  a  unit  of  surface,  and  the  area  of  the  rect- 
angle is  expressed  by  the  number  of  these  squares,  which  is  evidently  equal  to 
the  number  in  the  row  on  AB,  multiplied  by  tlie  number  of  such  rows,  or  the 
number  of  linear  units  in  AB  multiplied  by  the  number  in  AC'i 

If  the  two  sides  of  the  rectangle  are  not  commensurable,  take  some  very 
small  unit  of  length  which  will  divide  one  of  the  sides,  as  AC,  and  divide  the 
rectangle  into  squares  as  before ;  the  number  of  these  squares  will  be  the 
measure  of  the  rectangle,  except  a  small  part  along  one  side,  not  covered  by  the 
squares.  By  taking  a  still  smaller  unit,  the  part  left  unmeasured  by  the  squares 
will  be  still  less,  and  by  diminishing  the  unit  of  length  E,  we  can  make  the 
part  unmeasured  as  small  as  we  choose.  It  may,  therefore,  be  made  infinitely 
small  by  regarding  the  unit  of  measure  as  infinitesimal,  and  consequently  is  to 
be  neglected.*  Plence,  in  any  case,  the  area  of  a  rectangle  is  equal  to  the  pro- 
duct of  its  base  into  its  altitude,    q.  e.  d. 

321,  Cor.  1. —  The  area  of  a  square  is  equal  to  the  secoiid  poicer 
of  one  of  its  sides,  as  in  this  case  the  base  and  altitude  are  equal. 

822.  Cor.  2. — The  area  of  any  parallelogram  i.^  equal  to  the  pro- 
duct  of  its  base  into  its  altitude;  for  any  parallelogram  is  equivalent 
to  a  rectangle  of  the  same  base  and  altitude  {'J  13). 

823,  Cor.  3. — The  area  of  a  tria7igle  is  equal  to  one-half  the  pro- 
duct of  its  base  and  altitude;  for  a  triangle  is  one-half  of  a  parallelo- 
gram of  the  same  base  and  altitude  {314). 

324,  Cor.  4. — Parallelograms  or  triangles^  of  equal  bases  are  to 
each  other  as  their  altitudes ;  of  equal  altitudes,  as  their  bases ;  and 
in  general  they  are  to  each  other  as  the  products  of  their  bases  by 
their  altitudes. 


PROPOSITION  VII. 

825 •  Theorem, — The  area  of  a  traptezoid  xs  equal  to  the  product 
of  its  altitude  into  one-half  the  sum  of  its  parallel  sides,  or,  what  is 
the  same  thing,  tlie  product  of  its  altitude  and  a  line  joining  the 
middle  points  of  its  inclined  sides. 

*  This  principle  may  be  thus  stated:  An  infinitesimal  is  a  quantity  conceived,  and  to 
be  treated,  as  less  than  any  assignable  quantity ;  hence,  as  added  to  or  subtracted  from  finite 

quantities,  it  has  no  value.    Thus,  suppose  —  =  a,  w,  n,  and  a  bein;?  finite  quantities.    Let  c 

represent  an  infinitesimal;  then  — — -,  or  — ^,  or  "^^n.",  i^  to  be  considered  as  still  equal  to 
n  n  ±  c         n  r  c 

a,  for  to  consider  it  to  differ  from  a  by  any  amount  we  might  name,  would  be  to  assign  som4 

value  to  c. 

t  By  this  is  meant  the  areas  of  the  figures. 


142 


ELEMENTARY  PLANE   GEOMETRY. 


71   B 


Dexi.— In  the  trapezoid  ABCD   draw   either  diagonal,  as  AC.     It  is  thus 

divided  into  two  triangles,  whose  areas  are  to- 
gether equal  to  one- half  the  product  of  their 
common  altitude  (the  altitude  of  the  trapezoid), 
into  their  bases  DC  and  AB,or  this  altitude  into 
i  (AB  +  DC). 

Secondly,  if  ab  be  drawn  bisecting  AD  and 
CB,  then  is  «J  =  ^  (AB  +  CD).  For,  through 
a  and  h  draw  the  perpendiculars  om  and  pn^ 
meeting  DC  produced  when  necessary.  Now,  the  triangles  aoD  and  iKam  are 
equal,  since  Aa  =  aD,  angle  o  =  m,  both  being  right,  and  angle  oaD  =  Aam 
being  opposite.  Whence  Am  =  oD.  In  like  manner  we  may  show  that  Cp  = 
nB.  Hence,  ab  —  \{pp  +  mn)  =  i(AB  +  DC);  and  area  ABCD,  which  equals 
altitude  into  i(AB  +  DC),  =  aliitucU  into  ab.    q.  E.  d. 


Fig.  228. 


PROPOSITION  vm. 

326.  Theorem, — Tlie  area  of  a  regiilar  polygon  is  equal  to  one- 
half  the  product  of  its  apothem  into  its  perimeter. 


Dem. 


Let  ABCDEFG  be  a  regular  polygon  whose  apothem  is  Oa\  then  is 
its  area  equal  to  i  Oa  (AB  +  BC  +  CD  +  DE  +  EF 
+  PC  +  GA). 

Drawing  the  inscribed  circle,  the  radii  Oa,  Ob, 
etc.,  to  the  points  of  tangency,  and  the  radii  of  the 
circumscribed  circle  OA,  OB,  etc.  {264,  205),  the 
polygon  is  divided  into  as  many  equal  triangles  as 
it  has  sides.  Now,  the  apothem  (or  radius  of  the 
inscribed  circle)  is  the  common  altitude  of  these  tri- 
angles, and  their  bases  make  up  the  perimeter  of  the 
polygon.  Hence,  the  area  =  |Oa(AB  +  BC  +  CD 
4-  DE  +  EF  -f-  PC  +  CA).    Q.  E.  D. 


B       3       C 

Fig.  229. 


327,  Cor. — The  area  of  any  polygon  in  which  a  circle  can  le 
inscribed  is  equal  to  one-half  the  product  of  the  radius  of  the  in- 
scribed circle  into  the  perimeter. 

The  student  should  draw  a  figure  and  observe  the  fact.  It  is  especially 
worthy  of  note  in  the  case  of  a  triangle.    See  Fig.  60. 


PROPOSITION  IX. 

328.  Theorem. — Tlie  area  of  a  circle  is  equal  to  one-half  the 
product  of  its  radius  into  its  circumference. 


OF  AKEA. 


143 


Dem. — Let  Oa  be  the  radius  of  a  circle.  Circum- 
scribe any  regular  polygon.  Now  the  area  of  this 
polygon  is  one-half  the  product  of  its  apothem  and 
perimeter.  Conceive  the  number  of  sides  of  the 
polygon;  indefinitely  increased,  the  polygon  still 
continuing  to  be  circumscribed.  The  apothem  con- 
tinues to  be  the  radius  of  the  circle,  and  the  perim- 
eter approaches  the  circumference.  When,  there- 
fore, the  number  of  sides  of  the  polygon  becomes  in- 
finite, it  is  to  be  considered  as  coinciding  with  the  cir- 
cle, and  its  perimeter  with  the  circumference.  Hence 
the  area  of  the  circle  is  equal  to  one-half  the  pro- 
duct of  its  radius  into  its  circumference,     q.  e.  d. 

329,  Def. — A  Sector  is  a  part  of  a  circle  included  between  two 
radii  and  their  intercepted  arc.  Similar  Sectors  are  sectors  in  differ- 
ent circles,  which  have  equal  angles  at  the  centre. 

330,  Cor.  1. — The  area  of  a  sector  is  eqital  to  one-half  the  product 
of  the  radius  into  the  arc  of  the  sector. 

331,  Cor.  2. — The  area  of  a  sector  is  to  the  area  of  the  circle  r?.5 
the  arc  of  the  sector  is  to  the  circumference^  or  as  the  angle  of  the 
sector  is  to  four  right  angles. 


EXERCISES. 

1.  What  is  the  area  in  acres  of  a  triangle  whose  base  is  To  rods 
and  altitude  110  rods? 

2.  What  is  the  area  of  a  right  angled  triangle  whose  sides  about 
the  right  angle  are  126  feet  and  72  feet? 

3.  If  2  lines  be  drawn  from  the  vertex  of  a  triangle  to  the  base, 
dividing  the  base  into  parts  which  are  to  each  other  as  2,  3,  and  5, 
how  is  the  triangle  divided  ?  How  does  a  line  drawn  from  an  angle 
to  the  middle  of  the  opposite  side  divide  a  triangle  ? 

4.  Review  the  exercises  on  pages  49  and  50,  giving  the  reasons,  in 
each  case. 


lU 


ELEMENTARY  PLANE  GEOMETRY. 


SYNOPSIS. 


pa 
o 

M 


f  Definition. 

I  Prop.  I.  Of  parallelograms.  •{  Cor.  Paral.  and  rectangle. 

■r.  rr    r\P  *  '       i       i  Coj'.  1.  Triangle  and  rectangle. 

I  Prop.  II.  Of  triangles.  ^  ^^^    ^    Of  equal  bases  and  equaUltitudea. 

Prop.  III.  Square  on  i,  ^,  i  a  line,  etc. 

Prop.  IV.  Trapezoid. 

Prop.  V.  To  reduce  a  polygon  to  a  triangle. 


f  Cor.  1.  Of  square. 

Cor.  2.  Any  parallelogram. 

Prop.  VI.  Of  rectangle,  i   Cor.  3.  Of  triangle. 

I   Coi:  4.  Relation  of  parallelograms  and 
[  of  triangles. 


Prop.  VII.  Of  trapezoid. 

Prop.  VIII.  Of  regular  polygons. 


Cor.  Of  any  circumscribed 
polygon. 


Prop.  IX.  Of  a  circle. 


Def.  Of  sector. 

Cor.  1.  Area  of  sector. 

Cor.  2.  Relation  of  sector  to  circle. 


Exercises. 


SECTION  X, 

OF    SIMILARITY. 

55^.  The  primary  notion  of  similarity  is  UJceness  of  form.  Two 
figures  are  said  to  be  similar  which  hav-e  the  same  shape,  although 
they  may  differ  in  magnitude.*  A  more  scientific  definition  is  as 
follows : 

333,  Shnilar  Ftr/ures  are  such  as  hare  their  angles  respec- 
tively equal,  and  their  homologous  sides  proportional. 

334.  Homoloffous  Sides  of  similar  figures  are  those  which 
are  included  between  equal  angles  in  the  respective  figures. 


*  The  student  phould  be  careful,  at  the  outset,  to  mark  the  fact  that  nirniarify  involves 
Itco  things,  equalitt  op  angles  and  proportionalitt  op  sides.  It  will  appear  that,  in  the 
case  of  triangles,  if  one  of  these  facts  exists,  the  other  does  also ;  but  this  is  not  so  in  other 
polygons,  as  is  illustrated  in  Part  I. 


OF  SIMILARITY. 


U5 


In  Similar  Triangles,  the  Homologous   Sides  are  those 
opposite  the  equal  angles. 


PROPOSITION  I. 

33S.   Theorem, — Triangles  luliich  are  mutually  equiangular 
.are  similar. 


Dem.— Let  ABC  and  DEF  be  two  mu- 
tually equiangular  triangles,  in  which 
A=D,  B=E,  and  C  =  F;  then  are  the 
sides  opposite  these  equal  angles  propor- 
tional, and  the  triangles  possess  both 
requisites  of  similar  figures ;  ^.  e.,  they 
are  mutually  equiangular  and  have  their 
homologous  sides  proportional,  and  are 
consequently  similar. 

To  prove  that  the  sides  opposite  the 
equal  angles  are  proportional,  place  the 
triangle  DEF  upon  ABC,  so  that  F  shall 
coincide  with  its  equal  C,  CE'=FE,  and 
CD'=FD.  Draw  AE',  and  D'B.  Since  angle  CE'D'=CBA,  D'E'  is  parallel  to 
AB, and  as  the  triangles  D'E'A  and  D'E'B  have  a  common  base  D'E'  and  the 
same  altitude,  their  vertices  lying  in  a  line  parallel  to  their  base,  they  are 
equivalent  {316),  Now,  the  triangles  CD'E'  and  D'E'A,  having  a  common  alti- 
tude, are  to  each  other  as  their  bases  (324).    Hence, 


Fig.  231. 


For  like  reason 


CD'E'  :  D'E'A  :  :  CD'  :  D'A. 
CD'E'  :  D'E'B  :  :  CE'  :  E'B. 


Then,  since  D'E'A  and  D'E'B  are  equivalent,  the  two  proportions  have  a  com- 
mon ratio,  and  we  may  write  CD'  :  D'A  :  :  CE'  :  E'B. 

By  composition  CD' :  CD' +  D'A  :  :  CE'  :  CE'  +  E'B, 

or  CD'  :  CA  :  :  CE'  :  CB,     or  FD  :  CA  :  :  FE  :  CB. 

In  a  similar  manner,  by  applying  angle  E  to  B,  we  can  show  that 

I^E  :  CB  :  :  ED  :  BA.     Therefore,  FD  :  CA  :  :  FE  :  CB  :  :  ED  :  BA.     Q.  E.  D. 

S3G,  CoR.  1. — If  two  triangles  have  tiuo  angles  of  the  one  respec- 
tively equal  to  two  angles  of  the  other,  the  third  angles  heing  equal 
(221),  the  triangles  are  similar. 

337,  CoR.  2. — A  line  drawn  through  a  triangle  parallel  to  any 
side  divides  the  other  sides  proportionally. 

Thus  D'E'  being  parallel  to  AB,  it  is  shown  in  the  proposition  that 
CD':  D'A  ::CE':E'B. 

10 


146 


ELEMENTAEY  PLANE  GEOMETRY. 


338,  CoE.  3. — If  any  two  lines  cut  a  series  of  paraUels,  they  are 

divided  proportionally, 

O,  /O'  Dem. — If  the  two  'secant 

lines  are  parallel,  as  OA  and 
O'B',  the  intercepted  parts 
are  equal,  i.  e.^  ac  =  hd,  ee 
=  dL  eg  =  fh,  etc.  {24=2). 
Hence,  ac  :  bd  ::  ce  :  df  :: 
eg  :  fh.  Secondly,  if  the 
secant  lines  are  not  parallel, 
let  them  meet  in  some  point, 
as  O.  Then,  by  the  propo 
sition,  we  have 

Oa  :  ac  ::  Ob  :  bd    (1),        and  also  Oc  :  ce  ::  Od  :  df    (2). 

Taking  the  first  by  composition,  it  becomes 

Oa  +  ac  :  ac  : :  Ob  +  bd  :  bd,  or  Oc  :  ac  ::  Od  :  bd    (3). 

Now,  as  the  antecedents  in  (2)  and  (3)  are  the  same,  we  have 

ac  :  bd  ::  ce  :  df,  ot  ac  :  ce  ::  bd  :  df. 

In  like  manner,  we  may  show  that 

ce  :  df  :  :  eg  :  fh,  or  ce  :  eg  ::  df  :  fh. 


PROPOSITION  n. 

339,  T7ieorein,—Con\eYse\y,  If  tiuo  triangles  have  their  cor- 
responding sides  p)roportional,  they  are  similar. 

Dem.— In  the  triangles  ABC  and  DFE,  let  FD  :  CA  : :  FE  :  CB  :  :  DE  :  AB  ; 
then  are  the  triangles  similar. 

As  one  of  the  chai*acteristics  of  simi- 
larity, viz.,  proportionality  of  sides,  exists 
by  hypc-thesis,  we  have  only  to  prove 
the  other,  i.  e.,  that  the  triangles  are  mu- 
tually equiangular.  Make  CD'  equal  toFD, 
and  draw  D'E'  parallel  to  AB.  By  the 
preceding  proposition  CD'  (=  FD) :  CA  : : 
D'E'  :  AB.  But,  by  hypotliesis,  FD  : 
CA  ::  DE  :  AB.  Whence,  D'E'  =  DE. 
In  like  manner  CE'  :  CB  :  :  CD'  (=FD) : 
CA.  But,  by  hypothesis,  FE  :  CB  :  :  FD  : 
CA.  Whence  CE'  =  FE  ;  and  the  tiian- 
gle  CD'E'  is  equal  to  the  triangle  FDE 
{292).  Now,  CD'E'  and  CAB  are  mutu- 
ally equiangular,  since  D'E'  is  parallel  to  AB  (152)^  and  C  is  common.    Hence. 


Fig.  2.33. 


OF  SIMILARITY. 


147 


the  triangles  ABC  and  DEF  are  mutually  equiangular,  and  consequently  similar. 

Q.  E.    D. 

34:0,  Sen. — As  we  now  know  that  if  two  triangles  are  mutually  equiangular, 
they  are  similar;  or,  if  they  have  their  sides  proportional,  they  are  similar,  it  will 
be  sufficient  hereafter,  in  any  given  case,  to  prove  either  one  of.  these  facts,  in  order 
to  establish  the  similarity  of  two  triangles.  For,  either  fact  being  proved,  the 
other  follows  as  a  consequence.  See  Section  VI.,  Part  I.,  for  familiar  illustra- 
tions of  this  most  important  subject.  *  — 


PROPOSITION  III. 

34:1,  Theorem, — Tioo  triangles  which  have  the  sides  of  the  one 
respectively  2^(tralld  or  perpendicular  to  the  sides  of  the  other,  are 
similar. 

Dem.— Let  ABC  and  A'B'C  be  two  triangles  whose  sides  are  respectively 
parallel  or  perpendicular  to  each  other, 
then  are  the  triangles  similar. 

For,  any  angle  in  one  triangle  is 
either  equal  or  supplementary  to  the 
angle  in  the  other  which  is  included 
between  the  sides  which  are  parallel  or 
perpendicular  to  its  own  sides.  Thus  A 
either  equals  A',  or  A  +  A'  =  2  right 
angles  {281,  282,  283).  Now,  if  the 
corresponding  angles  are  all  supplemen- 
taiy,  that  is,  if  A  +  A'  =  2  R.A.,  B  +  B' 
:=2R.A.,andC  +  C  =  2  R.A.,  the  sum 
of  the  angles  of  the  two  triangles  is  6 
right  angles,  which  is  impossible.  Again, 
if  one  angle  in  one  triangle  equals  the 
corresponding  angle  in  the  other,  as  A 
=  A',  and  the  other  angles  are  supple- 
mentary, the  sum  is  4  right  angles  plus 
twice  the  equal  angle,  which  is  impossible.  Hence,  two  of  the  angles  of  one 
^triangle  must  be  equal  respectively  to  two  angles  of  the  other ;  and,  if  two  are 
equal,  the  third  angle  in  one  is  equal  to  the  third  in  the  other  {221).  Hence, 
the  tiiangles  are  mutually  equiangular,  and  therefore  similar  {335).    Q.  e.  d. 


PROPOSITION  IV. 

"^42,  OrJieorem. — Two  triangles,  which  have  an  angle  in  each 
equal,  and  the  sides  about  the  equal  a7igles proportional,  are  similat. 


148 


ELEMENTAEY  PLANE   GEOMETRY. 


Fig.  235. 


Dem.— In  the  triangles  ABC  and  DEF 
let  C  =  F,  and  AC  :  DF  ::  CB  :  FE; 
then  are  the  triangles  similar. 

For,  place  F  on  its  equal  C,  and  let  D 
fall  at  D'.  Draw  D'E'  parallel  to  AB. 
Then  AC  :  D'C  (=  DF)  ::  BC  :  CE'  (337). 
But  by  hypothesis  AC  :  DF  : :  BC  :  FE. 
.'.  CE'  =  FE,  and  the  triangles  D'CE'  and 
DFE  are  equal  (284).  Therefore,  D'CE' 
being  equiangular  with  ACB,  is  similai 
to  it  (835) ;  and  as  DFE  is  equal  to  D'CE', 
DFE  is  similar  to  ACB.    q.  e.  d. 


PROPOSITION  y. 

343.  Theorem, — In  any  right  angled  triangle,  if  a  line  he 
drawn  from  the  right  angle  perpendicular  to  the  hypotenuse,  it 
divides  the  triangle  into  two  triangles,  which  are  similar  to  the  given 
triangle,  and  consequently  similar  to  each  other. 

Dem.— Let  ACB  be  a  triangle  right-angled  at  C,  and  CD  a  perpendicular 
upon  the  hypotenuse  AB ;  then  are  ACD  and  CDB 
similar  to  ACB,  and  consequently  to  each  other. 

For,  the  triangles  ACD  and  ACB  have  the  angle  A 
common,  and  a  right  angle  in  each;  hence  they  are 
mutually  equiangular,  and  consequently  similar(556). 
For  a  like  reason  CDB  and  ACB  are  similar.  Finally, 
as  ACD  and  CDB  are  both  similar  to  ACB,  they  are 

Q.   E.   D. 


A  D       B 

Fig.  236. 

similar  to  each  other. 


344,  Cor.  1. — Either  side  about  the  right  angle  is  a  mean  propor- 
tional letween  the  wJiole  hypotenuse  a7id  the  adjacent  segment. 

Dem. — This  is  a  direct  consequence  of  the  similarity  of  the  partial  triangles 
with  the  whole  trinngle.  Thus,  comparing  the  homologous  sides  of  ACD 
and  ACB,  we  have  AD  :  AC  : :  AC  :  AB  ;*  and  from  CDB  and  ACB,  we  have 
DB  :  CB  ::  CB  :  AB. 

34o.  Cor.  2. — The  perpendicular  is  a  mean  proportional  between 
the  segments  of  the  hypotenuse. 

Dem. — This  is  a  consequence  of  the  similarity  of  ACD  and  CDB.  Thus, 
AD  :  CD  :  :  CD  :  DB. 

*  Notice  that  AD  of  the  triangle  ACD  is  opposite  angle  ACD,  and  AC,  i^8  consequent,  is 
of  the  triangle  ACB,  and  opposite  the  angle  B,  which  equals  angle  ACD-  The  student  must 
be  sure  that  he  knows  in  what  order  to  take  the  sides,  and  ivhy. 


OF  SIMILARITY.  149 

Queries.— -To  -which  triangle  does  the  fii-st  CD  belong  ?  To  which  the  second  ? 
Why  is  CD  made  the  consequent  of  AD  ?  Why,  in  the  second  ratio,  are  CD  and 
DB  to  be  compared  ? 

346,  Cor.  3. — The  square  described  on  the  hypotenuse  of  a  right 
angled  triangle  is  equivalent  to  the  sum  of  the  squares  described  on 
the  other  tiuo  sides. 

Dem.— From  Ow.  1,  AC*  =r  AB  x  AD 

and  also  CB''  =  ABxDB. 

Therefore,  adding,  AC'  +  CB'  =AB  (AD  +  DB)  =  AB". 

547.  Cor.  4. — If  a  perpe^idicular  be  let  fall  from  any  point  in  a 
circumference  upon  a  diameter,  this  perpendic- 
ular is  a  mean  ptroportional  between  the  seg- 
ments of  the  diameter. 

Dem.— Thus,  AD  :  CD  : :  CD  :  DB,  or  CD'  =:AD  x  DB. 

For,  drawing  AC  and  CB,  ACB  is  a  right  angle, 
and  the  case  falls  under  Cor.  2. 

The  chords  AC  and  CB  are  mean  proportionals  between  the  whole  diameter 
and  their  adjacent  segments  by  Cor.  1. 

34:8,  ScH.— This  proposition,  with  its  corollaries,  is  perhaps  the  most  fruit- 
ful in  direct  practical  results  of  any  in  Geometiy.  Coi'.  3  will  be  recognized 
as  a  demonstration  of  the  Pj-thagorean  proposition  {109),  Part  I.  There  are 
many  other  demonstrations  of  exceeding  beauty,  some  of  which  will  be  given 
in  Part  III.  The  one  here  given  is  the  simplest,  and  shows  best  the  way  in 
which  this  truth  grows  out  of  the  more  general  fact  of  similarity. 


PROPOSITION  VI. 


w 


340,  Tlieorein, — Regular  p)olygons  of  the  same  number  of  sides 
are  similar  figures. 

Dem. — Let  P  and  P'  be  two  regular  polygons  of  the  same  number  of  sides,* 
a,  5,  c,  d,  etc.,  being  the  sides  of  the  former,  and  a',  b\  c',  d\  etc.,  the  sides  of 
the  latter.  Now,  by  the  definition  of  regular  polygons,  the  sides  «,  b,  c,  d, 
etc.,  are  equal  each  to  each,  and  also  a\  b\  c\  d\  etc.  Hence,  we  have 
a  :a'  :  :b  :b'  :  '.  e:  c'  :  :  d:  d',  etc.  Again,  the  angles  are  equal,  since  n  being 
the  number  of  sides  of  each  polygon,  each  angle  is 

71  X  2  right  angles  —  4  right  angles     ^^-^/^n 
n 
Hence  the  pol5''gons  are  mutually  equiangular,  and  have  their  sides  proportional ; 
that  is,  they  are  similar,     q.  e.  d. 

*  The  etudent  may  conetnict  two  regular  hexagons,  if  thought  desirable. 


150 


ELEMENTARY  PLANE  GEOMETRY. 


350.  Cor.  1. — Tlie  corresponding  diagonals  of  regular  polygons 
of  the  same  number  of  sides  are  in  the  same  ratio  as  the  sides  of  the 
polygons. 

Let  the  student  draw  a  figure  and  demonstrate  the  fact. 

351.  Cor.  2. — The  radii  of  the  inscribed,  and  also  of  the  circuni- 
P^^_^^g^  scribed  circles,  of  regular  polygons  of  the 

same  nxiniber  of  sides,  are  in  the  same  ratio 
as  the  sides  of  the  polygons. 

Dem. — Since  the  angles  F  and/  are  equal,  and 
bisected  by  Fd  the  right  angled  triangles  OSF, 
0^/  are  equiangular,  and  hence  similar.  There- 
fore FS  :  /5  :  :  SO  :  «0  or  FO  :  /O.  Whence, 
doubling  both  terms  of  the  first  couplet, 
Fig.  2^.  FA  :  /a  :  :  SO   :  sO  or  FO  :  /O. 


PROPOSITION  ^TL 

■Circles  are  similar  figures. 

Dem. — Let  Oa  and  OA    be    the   radii    of  any 
two  cii'cles.     Place  the   circles  so  that  they  shall 
be  concentric,  as  in  the  figure.     Inscribe  the  regu- 
lar hexagons,  as  abcdef,  ABCDEF.     Conceive  the 
arcs  AB,  BC,  etc.,  of  the  outer  circumference,  bi- 
sected, and  the  regular  dodecagon  inscribed,  and 
also  tiie  corresponding  regular  dodecagon  in  the 
inner  circumference.      These  are  similar  figures 
hj  {34:9).    Now,  as  the  process  of  bisecting  the 
arcs  of  the  exterior  circumference  can  be  con- 
ceived as  indefinitely  repeated,  and  the  coiTespouding  regular  polygons  as  in- 
scribed in  each  circle,  the  circles  may  be  considered  as  regular  polygons  of  the 
same  number  of  sides,  and  hence  similar,    q.  e.  d. 

353.  Cor. — Arcs  of  similar  sectors  are  to  each  other  as  the  radii 
of  their  circles;  i.  e.,  arc /e  :  arc  FE  :  :  0/:  OF. 

ScH. — The  circle  is  said  to  be  the  Limit  of  the  inscribed  polygon,  and 
the  circumference  the  limit  of  the  perimeter.  By  this  is  meant  that  as  the 
number  of  the  sides  of  the  inscribed  polygon  is  increased  it  approaches  nearer 
and  nearer  to  equality  with  the  circle.  The  apothem  approaches  equality  with 
the  radius,  and  hence  has  the  radius  for  its  limit.  It  is  an  axiom  of  great 
importance  in  mathematics  that,  Whatever  can  he  sfiown  to  he  true  of  a  magni- 
iude  as  it  approaches  its  limit  indefinitely,  is  true  of  that  limit. 


OF  SIMILARITY. 


151 


ABC 


B   A 


EXERCISES. 

1.  l^Toh, — To  divide  a  given  line  into  ixirts 
which  shall  le  proportional  to  several  given 
lines.   ' 

Solution. — Let  it  be  required  to  divide  OP  into 
parts  proportional  to  tlie  lines  A,  B,  C,  and  D.  Draw 
ON  making  any  convenient  angle  with  OP,  and  on  it 
lay  off  A,  B,  C,  and  D,  in  succession,  terminating  at 
M.  Join  M  with  the  extremity  P,  and  draw  parallels  to 
MP  through  the  other  points  of  division.  Then  by 
reason  of  the  parallels  we  shall  have 

A:B:C:D::«:6:c:d,    {338). 

2.  I^rob. —  To  find  a  fourth  proportional  to  three  given  lines. 
For  the  solution  see  {89).    Repeat  the  process,  and  give  the  reasons. 

3.  I^rob, — To  find  a  third  propor- 
tional to  tiuo  given  lines. 

Solution. — This  may  be  solved  as  the  two 
preceding.  Thus,  take  any  two  lines,  as  A  and 
B,  for  the  given  lines.  We  are  to  find  a  third 
line  X,  such  that  A  :  B  : -.  B  .  x.  The  figure 
will  suggest  the  details. 

The  following  is  a  solution  based  on  {347). 
Draw  an  indefinite  line  AM.  Take  AD  =  A, 
and  erect  BD  =  B.  Join  AB,  and  bisect  it  by 
the  perpendicular  ON.  Then  with  O  as  a  cen- 
tre, and  OA  as  a  radius,  describe  a'semi-circum-  _ 
ference.  This  will  pass  through  B.  (Why?)  M 
Also  AD  :  BD  ::  BD   :  CD  (=  x).     (Why?) 

4.  Draw  any  straight  line  on  the  blackboard,  and  divide  it  into  5 
equal  parts,  upon  the  principle  used  in  the  preceding  solutions. 

5.  Review  the  exercises  under  (89,  90),  and  give  the  reasons. 

6.  IProh, — To  find  a  mean  proportioiial  betiveen  two  given  lines. 

For  the  solution  see  {110).    Repeat  the  process,  and  give  q 

the  reasons  for  the  method. 

• 

7.  DE  being  parallel  to  BO,  prove  that  the  triangles 
DOE  and  BOG  are  similar,  and  hence  that  OD  :  00  :: 
OE  :  OB.    Are  the  following  proportions  true? 

OD   :  00  ::  OE   :  OB,     OD   :  DE  : :  00  :  BC, 
OD   :  OE  ::  00   :  OB,     OB    :  BC  : :  OE   :  DE. 


O     D 


Fig.  Si4\ 


152 


ELEMENTARY  PLANE  GEOMETRY. 


8,  Show  that  if  ABCDEF  is  a  regular  polygon,  kbcdef  is  also  regular, 
Ic,  cd,  etc.,  being  parallel  to  BC,  CD,  etc. 
Show  that  any  two  similar  polygons  may  be 
placed  in  similar  relative  positions,  and 
hence  show  that  the  corresponding  diagonals 
are  in  the  same  ratio  as  the  homologous 
sides. 


Fig.  iM3. 

homologous  with 
latter  ? 


9.  The  sides  of  one  triangle  are  7,  9,  and 

11.    The  side  of  a  second  similar  triangle, 

side  9,  is  4J.    What  are  the  other  sides  of  the 


10.  The  diameter  of  a  circle  is  20.  What  is  the  perpendicular 
distance  to  the  circumference  from  a  point  in  the  diameter  15  from 
one  extremity  ?  What  are  the  distances  from  the  point  where  this 
perpendicular  meets  the  circumference  to  the  extremities  of  the 
diameter? 


SYNOPSIS. 


O 


Primary  notion  of  similarity. 
Definition  of  similarity. 
Homogeneity  of  sides.    In  general. 


5    r 


S3 

fa  5 


S 
§ 


Prop.  L  Mutually  equiangular 


In  triangles. 

{Cor.  1.  Two  angles  equal. 
Cor.  3.  A  parallel  to  a  side. 
Cor.  3.  Lines  cutting  parallels. 

■I  Sell.  Either  of  two  facts  sufficient 


Prop.  II.  Sides  proportional. 

Prop.  III.  Sides  parallel  or  perpendicular. 

Prop.  IV.  An  angle  equal  in  each,  and  sides  proportional. 

'  Cor.  1.  Side  about  right  angle. 


Prop.  V.  Perpendicular  from 
right  angle  upon 
hypotenuse. 


Prop.  VI.  Regular  polygons  similar. 
Prop.  VII.  Circles  similar. 


Cor.  2.  Perpendicular, 
Cor.  3.  Square  on  hypotenuse. 
Cor.  4.  Perpendicular  on  diameter. 
Sch.  Importance  of  this  Prop,  and 
Cor's. 

Cor.  1.  Corresponding  diagonals. 
Cor.  2.  Radii  of  inscribed  and  cir- 
cumscribed circles. 

Sch.  Circle  limit  of  polygon. 


Exercises. 


Prob.  To  divide  a  line  into  proportional  parts. 
Prob.  To  find  a  fourth  proportional. 
Prob.  To  find  a  third  proportional. 
Prob.  To  find  u  mean  proportiouiil. 


APPLICATIONS   OF  THE  DOCTRINE  OF   SIMILARITY.  153 


SECTION  XL 

APPLICATIONS    OF    THE    DOCTRINE    OF    SIMILARITY    TO    THE 
DEVELOPMENT  OF  GEOMETRICAL  PROPERTIES  OF  FIGURES. 


354:,  The  doctrine  of  similarity,  as  presented  in  the  preceding 
section,  is  the  chief  reliance  for  the  development  of  the  geometrical 
properties  of  figures.  This  section  will  be  devoted  to  the  investiga- 
tion of  a  few  of  the  more  elementary  properties  of  plane  figures, 
which  we  are  able  to  discover  by  means  of  this  doctrine. 


OF  THE  RELATIONS  OF  THE  SEGMENTS  OF  TWO  LINES  INTERSECT- 
ING EACH  OTHER,  AND  INTERSECTED  BY  A  CIRCUMFERENCE. 


PROPOSITION  I. 

355.  Tlieovenu — If  tioo  cliords  intersect  each  other  in  a  circle, 
their  segments  are  reciprocally  j)v02)ortional ;  whence  the  product  of 
the  segments  of  one  chord  equals  the  2)roduct  of  the  segments  of  the 
other. 

Dem.— Let  the  chords  AC  and  BD  inter8ect  at  0  ;  then  is  AG  :  BO  : :  DO  : 
CO,  whence  AO  x  CO  ==  BO  x  DO. 

For,  draw  AD  and  BC.  The  two  triangles  AOD  and  BOC  ^^'Jt~^\ 

are  equiangular,  and  hence  similar ;  since  the  angles  at  O         /^  /"\     J^ 
are  vertical,  and  consequently  equal  {134),  and  D  =  C,      /      /      ^ 
because  both  are  measured  by  i  arc  AB  {210).    (A  =  B     /     /  /^^ 
because  both  are  measured  by  i  arc  DC  ;  but  it  is  only     I  /y^         ^ 
necessary  to  show  that  two  angles  are  equal  in  order  to     }\ 
show  that  the  triangles  are  equiangular,  and  hence  simi-       \ 

lar.)    Now,  comparing  the  homologous  sides  (those  oppo-  ^ ^-^B 

site  the  equal  angles),  we  have  AO   :  BO   :  :  DO  :  CO ;  Fig.  244. 

whence,  AO  x  CO  =  BO  x  DO.     q.  e.  d. 

Queries.— Why  is  AO  compared  with  BO?  Why  DO  with  CO?  Would 
AO  :  CO  :  :  BO  :  DO  be  true?  Would  AO  :  DO  :  :  BO  :  CO?  What  is  the 
force  of  the  word  "reciprocally,"  as  used  in  the  proposition? 


154 


ELEMENTARY  PLANE  GEOMETRY. 


PROPOSITION  n. 

356.  TJieoreni, — If  from  a  point  without  a  circle,  two  seca7its 
he  draiun  terminating  in  the  concave  arc,  the  lohole  secants  are  recip- 
rocally proportional  to  their  external  segments;  ivhence  the  product 
of  one  secant  into  its  external  segment  equals  the  product  of  the  other 
Q^    into  its  external  segment. 

Dem. — OA  and  OB  being  secants,  OA  :  OB  :  : 
OC  :  OD,  and  consequentlj'  OA  x  OD  =  OB  x  OC. 
For,  drawing  AC  and  DB,  the  two  triangles  AGO 
and  BDO  have  angle  0  common,  and  A  =  B,  since 
both  are  measured  bj'  \  DC  ;  hence  the  triangles  are 
similar,  and  we  have  OA  :  OB  : :  OC  :  OD,  and 
consequently  OA  x  OD=;OB  x  OC.     q.  e.  d. 

Same  queries  as  under  the  preceding  demon stra- 
FiG.  245.  tion. 


PROPOSITION  in. 

^  357.  TJieorem. — If  from  a  point  without  a  circle  a  tangent  he 
dravm,  and  a  secant  terminating  in  the  concave  arc,  the  tangent  is  a 
mean  proportio?ial  hetween  the  whole  secant  and  its  external  seg- 
ment; whence  the  square  of  the  tangent  equals 
the  product  of  the  secant  into  its  external  seg- 
ment. 

Dem.— OA  being  a  tangent  and  OB  a  secant,  OB  : 
OA  :  :  OA  :  OC,  whence  OA^  =  OB  x  OC.  For, 
drawing  AB  and  AC,  the  triangles  OAB  and  ACO  have 
angle  O  common,  and  OAC  =  B,  since  each  is  measured 
by  i  arc  AC;  hence  the  triangles  are  similar,  and  OB  : 
OA  : :  OA  :  OC,  whence  OV  =  OB  x  OC.    q.  e.  d. 


Fig.  246. 


OF  THE  BISECTOR  OF  AN  ANGLE  OF  A  TRIANGLE. 


PROPOSITION   IV. 

358.  Tlieorein. — A  line  whicli  bisects  any  angle  of  a  triangle 
divides  the  opposite  side  into  segments  proportional  to  the  adjacent 
sides. 


BISECTOR  OF  AN  ANGLE  OP  A  TRIANGLE. 


155 


Dem.— Let  CD  bisect  the  angle  ACB;  then 
AD  :  DB  ; :  AC  :  CB. 

For,  draw  BE  parallel  to  CD,  and  produce  it 
till  it  meets  AC  produced,  as  at  E.  Now,  by- 
reason  of  the  parallels  CD  and  EB,  angle  ACD 
=  AEB,  and  angle  DCB  =  CBE  {152).  Whence, 
as  ACD  =  DCB  by  hypothesis,  E  =  CBE,  and 
CE  =  CB  {227)-  Also,  since  CD  is  parallel  to 
EB,  AD  :  DB  : :  AC  :  CE  {337).  Substituting 
for  CE  its  equal  CB,  we  have  AD  :  DB  : :  AC 


E.   D. 


PROPOSITION  Y. 

S50»  Tlieorenn, — If  a  line  le  draivn  from  any  vertex  of  a 
triangle  hisecting  the  exterior  a?igle  and  intersecting  the  ojjposite 
side  jyroduced,  the  distances  from  the  other  vertices  to  this  intersection 
are  lyrojportional  to  the  adjacent  sides. 

Dem.— Through  the  vertex  C  let  CD 
be  drawn,  bisecting  the  exterior  angle 
FCB,  and  intersecting  AB  produced  in 
D ;  then  AD  :  BD  : :  AC  :  CB. 

For,  draw  BE  parallel  to  CD.  By  rea- 
son of  these  parallels,  angle  FCB=:  CEB, 
andBQD=CBE(i5;?).  Hence  CEB  = 
CBE,  and  CB  =  CE.  Also,  by  reason  of 
the  parallels,   AD  :  BD  ::  AG  :  CE,   or  its  equal  CB  (555).    Q.  E.  d. 


PROPOSITION  yi. 

360,  TJieorem. — If  a  line  be  draiun  hisecting  any  angle  of  a 
triangle  and  intersecting  the  opposite  side,  the  rectangle  of  the  sides 
ahoiit  the  bisected  angle  equals  the  rectangle  of  the  segments  of  the 
third  side,pflf^s  the  square  of  the  bisector. 

Dem.— Let  CD  bisect  the  angle  ACB  ;  then  AC  x  CB 
=^  AD  X  DB  +  CD". 

For,  circumscribe  the  circle  about  the  triangle,  pro- 
duce the  bisector  till  it  meets  the  circumference  at  E, 
and  draw  EB.  The  triangles  ADC  and  CBE  are  simi- 
lar, since  angle  ACD  =  EQB,  by  hypothesis,  and  A  =  E, 
because  each  is  measured  by  i  arc  CB.  Therefore, 
AC  :  CE  : :  CD  :  CB,  whence  AC  x  CB  =  CE  x  CD 
=  (DE  +  CD)  CD  =  DE  x  CD  +  CD*.  For  DE  x 
CD,  substituting  its  equivalent  AD  x  DB  {355),  we 
have  AC  X  CB  =  AD  X  DB  +.  CD^    Q.  e.  d. 


-\B 


E. 
Fig.  a49. 


156  ELEMENTARY  PLANE  GEOMETRY. 


PROPOSITION  yn. 

361,  Theorem, — The  bisectors  of  the  angles  of  a  triangle  all 
pass  through  the  same  point,  which  is  the  centre  of  the  inscribed 
circle. 

Dem. — Draw  two  lines  bisecting  two  of  the  angles,  and  from  their  inter- 
section draw  a  line  to  the  other  angle.  Then  show  that  the  latter  angle  is 
bisected.  By  (Ex.  4,  page  134)  this  point  is  shown  to  be  the  centre  of  the  in- 
scribed circle.     [The  student  should  fill  out  the  demonstration.] 


AEEAS  OF  SIMILAR  FIGURES. 


PROPOSITION   Yin. 

362,  Theore'in, — Tlie  areas  of  similar  triangles  are  to  each 
other  as  the  squares  described  on  their  homologous  ^ides. 
Dem. — Let  ABC  and  DEF  be  any  two  similar  triangles ;  then  is 

area  ABC   :  area  DEF  : :  CB'  :  FE"  : :  AC '  :  DF'  :  •  AB"  :  DE' 

For,  place  the  largest  angle  of  the  ti'iangle 
DEF,  as  D,  on  its  equal  angle  A,  of  the  triangle 
ABC*;  let  E  fall  at  E',  F  at  F',  and  draw 
ET';  then  is  triangle  AE'F'  =  DEF  {284),  and 
E'F'  is  parallel  to  BC.  Let  fall  a  pei-pendicular 
from  A  to  CB.  Then  Al  is  the  altitude  of  AE'F', 
and  AH  of  ABC.  Now,  by  similar  triangles  we 
liave  CB  :  F'E'  :  :  AH  :  Al. 
But  iAH  :  iAI  :  :  AH  :  Al ;  and,  multiplying 
corresponding  terms,  ^  AH  x  CB  :  ^Ai  x  F'E'  : :  AH*:  AlV     Whence,  since 

•^AH  X  CB  =  area  ABC,  and  lAI  x  F'E'  =  area  AE'F'=  area  DEF,  and 
AH  :  Al  : :  CB  :  FE  :  :  AC  :  DF  :  :  AB  :  DE,  or  AH*  :"aI'  : :  CB'  :  FE'  : :  A'c'  :~DF" 


Q.  E.  D. 


PROPOSITION   IX. 
363,  Hieoreni, — The  areas  of  similar  polygons  are  to  each 
other  as  the  squares  of  the  homologous  sides  of  the  polygons. 

*  The  only  object  in  taking  the  largest  angle  is  to  make  the  perpendicular  fall  wUhin  the 
triangle.    Any  two  equal  angles  may  be  applied,  and  the  demonstration  is  essentially  the  same. 


^  :  Al: 

:CB 

FE 

::AC: 

DF:: 

AB:DE 

or 

ah'  :  Al' 

::CB' 

:FE' 

AB':: 

DE% 

we 

have 

area  ABC 

:  area  DEF 

::  CB 

^   FE^: 

AC' 

:   DF'  :  : 

AB': 

DE'. 

AREAS   OF  SIMILAR   FIGURES. 

Dem.— Let  ahcdef  and  ABCDEF 
be  two  similar  polygons.  Desig- 
nate the  former  by  p,  and  the  lat- 
ter by  P.  Then  p:  P  ::ab' ::  AB^* 
or  as  the  squares  of  any  other  two 
homologous  sides. 

For,  from  the  equal  angles  a 
and  A  drawing  the  diagonals,  the 
corresponding  partial  angles  into 
which  a  and    A    are    divided  are  yig.  2o1. 

equal.  [Let  the  student  show  why  hj  34:2^]  Now  take  A&'=«6,  and  draw 
I'c',  making  angle  A6'c'  =  h.  Then  h'c'  =  be,  and  Ac'  =  ac,  since  the  triangles 
abc  and  A&'c'  have  two  angles  and  the  included  side  of  one  equal  to  two  angles 
and  the  included  side  of  the  other.  In  like  manner  draw  c'd'  making  angle 
b'c'd'=  bed,  e'd'=  ed,  and  AcZ'=  ad.  So,  also,  making  angle  e'd'e'=  cde,  and  angle 
d'e'f  =  def,  die'  —  de,  e'f  —  ef,  and  /'A  =fa.  Hence  the  polygon  Nb'c'd'e'f'=p, 
and  its  sides  are  respectively  parallel  to  the  corresponding  sides  of  P.  Now,  let 
m,  n,  0,  and  s  represent  the  triangles  in  which  they  stand,  and  M,  N,  0,  and  S 
the  corresponding  triangles  of  P,  as  AFE,  etc.  Triangles  m  and  M  being  similar, 
and  also  n  and  N,  we  have 

m  :  M  :  Cf^^  :  AE",  and  ti  :  N  :  :  a7''  :  AE*. 
Whence  w  :  M  : :  ?t  :  N. 

In  like  manner  we  can  show  that  7i  :  N  :  :  o  :  0,  and  that  o  :  0  :  :  8  :S. 
Whence  m  :  M  :  :  n  -M  :  -  o :  0  :  :  s  :  S. 

By  composition,  (m  +  n  +  o  +  s)  (or  j9)  :  (M  +  N  +  0  f  S)  (or  P)  : :  «  :  S, 
But  «  :  S  :  ~M'^  {ov~ab^)  :'AB\    Therefore  p  :  P  :  :'ab^ :  AB\  or  as  the  squares  of 
any  two  homologous  sides.    Q.  e.  d. 

364,  CoR.  1. — Similar  polygons'^  are  to  each  other  as  the  squares 
of  their  c07'responding  diagonals. 

In  the  demonstration  we  have  « :  S  : :  Ac*  (or  ac) :  AC^  Whence  p  :  P  :  :  ac" :  AC*. 
The  same  may  be  shown  of  any  other  corresponding  diagonals. 

SOS,  Cor.  2. — Regular  polygons*  of  the  same  numler  of  sides 
are  to  each  other  as  the  squares  of  their  homologous  sides  ;  since  they 
are  similar  figures  (349). 

366,  Cor.  3. — Regular  polygons*  of  the  same  number  of  sides 
are  to  each  other  as  the  squares  of  their  apothems. 

For  their  apothems  are  to  each  other  as  their  sides.  Hence  the  squares  of 
their  apothems  are  to  each  other  as  the  squares  of  their  sides. 

367,  Cor.  4. — Circles  are  to  each  other  as  the  squares  of  their 
radii  {3d2),  and  as  the  squares  of  their  diameters. 


*  This  is  a  common  elliptical  form  for  "  The  areas  of,  etc." 


158  ELEMENTARY   PLANE   GEOMETRY. 


OF  PERIMETERS  AND   THE  RECTIFICATION  OF  THE 
CIRCUxMFERENCE. 

368,  Tlie  Rectification  of  a  curve  is  the  process  of  finding 
its  length. 

The  term  rectijication  signifies  making  straight,  and  is  applied  as  above, 
under  the  conception  that  the  process  consists  in  finding  a  straight  line  equal 
in  length  to  the  cuitc. 


PROPOSITION  X. 

369*  Hieorem, — The  perimeters  of  similar  polygons  are  to 
each  other  as  their  homologous  sides,  and  as  their  corresponding 
diagonals. 

Dem.— Let  a,  J,  c,  d,  etc.,  and  A,  B,  C,  D,  etc.,  be  the  homologous  sides 
of  two  similar  polygons  whose  perimeters  are  p  and  P ;  then  p  :  P  :  :  a  : 
A  :  :  b  :  B  : :  c  :  C,  etc. ;  and  r  and  R  being  corresponding  diagonals,  p  : 
P  : :  r  :  B.  Since  the  polygons  are  similar,  a  :  A  :  .  b  :  B  : :  c  :  C  : :  rf  :  D, 
etc.  By  composition,  {a  +  b  +  c  +  d+ etc.)  (or  ^) :  (A  +  B  +  C  +  D  +  etc.)  (or  P) :  :  a :  A, 
or  as  any  other  homologous  sides.  Also,  as  the  homologous  sides  are  to  each 
other  as  the  corresponding  diagonals  {350),  p  :  P  :  :  r  :  M.    Q.  e.  d. 

370,  Cor.  1. — The  perimeters  of  regular  polygons  of  the  same 
numder  of  sides  are  to  each  other  as  the  apothems  of  the  polygons. 

For  the  apothems  are  to  each  other  as  the  sides  of  the  polygons  {351). 

371,  Cor.  2. — Tlie  circumferences  of  circles  are  to  each  other  as 
their  radii,  and  as  their  diameters  ;  since  they  may  be  considered  as 
regular  polygons  of  the  same  number  of  sides  (352), 


PROPOSITION  XI. 

372,  Theorem, — The  circumference  of  a  circle  tvhose  radius  is 
1,  is  27r.  the  numerical  value  of  n  being  approximately  3.1416. 

Dem. — We  will  approximate  the  circumference  of  a 
ch-cle  whose  radius  is  1,  by  obtaining,  1st,  the  perim- 
eter of  the  regular  inscribed  hexagon ;  2d,  the  perim- 
eter of  the  regular  inscribed  dodecagon;  3d,  the 
perimeter  of  the  regular  inscribed  polygon  of  24  sides ; 
then  of  48,  etc. 

In  order  to  do  this,  let  us  find  the  relation  between 

the  chord  of  an   arc  and  the  chord  of  \  the  arc  in  a 

Fig.  252.  circle  whosc  radius  is  1.     Call  the  chord  of  an  arc  as 


RECTIFICATION   OF  CIRCUMFERENCE.  159 

AB,  (7,  and  the  cliorcl  of  half  the  arc,  as  CB,  c.  Now,  BDO  is  right  angled  at 
D,  whence "BO'  ="60'  +  DO'' (346),  or  DO  =  ^/bo'-BD".  But  in  the  present 
case  BO  =  1 ;  hence  DO  =  VT^W\  Taking  DO  from  CO,  we  have  CD  =  1  - 
-/T^fC*-  From  the  right  angled  triangle  BDC  we  have  CB  (or  c)  = 
-/bd"  +  W,  or  substituting  i  G  for  BD,  and  1  -  VT^^JC^  for  CD,  this  re- 
duces to  c=v'2^~P5z^^    or,     [2-(4-0'2)^]^ 

By  the  use  of  this  formula,  we  make  the  following  computations  : 
No.  sides.  Form  of  Compntetion.  Length  of  Side.  Perimeter. 

6.  ^ee{271) 1.00000000  6.00000000 

12.  e=^^Zvi^=  V2-  Vd,  or  (2-3')^  .        =  .51763809  6.21165708 

24.  c=  ■J2-[4-(2-3^)]^p  =  [2-(2  +  3^)']^       .        =  .26105238  6.26525722 

=  j  2-  [2  +  (2  +  8^)']^  }  ^  =  .13080626  6.27870041 

96.  c  =  (2-  -j  2  +  [2  +  (2  +  8')']^  \^y         .        .        .        =  .06543817  6.28206396 

192.  c  =  [2-(2+-J2  +  [2  +  (2  +  3^y]'p)*]'     .        .        =.03272346  6.28290510 

384.  c=  ]2-[2  +  (24-]2  +  [2  +  (2-f3'y]'p)']4'  =  -01636228  6.28311544 

768.  c  =  (2--{2  +  [2  +  (2+|2  +  [2  +  (2  +  3^)]4^")']4')'=  -00818121   6.28316941 

It  now  appears  that  the  first  four  decimal  figures  do  not  change  as  the  num- 
ber of  sides  is  increased,  but  will  remain  the  same  how  far  soever  we  proceed. 
We  may  therefore  consider  6.28317,  as  approximately  the  circumference  of  a 
circle  whose  radius  is  1,  i.  e.,  27r  =  6.28317,  nearly  ;  and  re  =  3.1416,  nearly. 

373.  Sen. — The  symbol  7t  is  much  used  in  mathematics,  and  signifies, 
primarily,  tJie  semi-circumference  of  a  circle  w/iose  radius  is  1.  ^tc  \s  therefore 
a  symbol  for  a  quadrant,  90°,  or  a  right  angle.  :|  ;r  is  equivalent  to  45°,  etc., 
the  radius  being  always  supposed  1,  unless  statement  is  made  to  the  contrary. 
The  numerical  value  of  7t  has  been  sought  in  a  great  variety  of  ways,  all  of 
which  agree  in  the  conclusion  that  it  cannot  be  exactly  expressed  in  decimal 
numbers,  but  is  approximately  as  given  in  the  proposition.  From  the  time  of 
Archimedes  (287  B.C.)  to  the  present,  much  ingenious  labor  has  beeu  bestowed 
upon  this  problem.  The  most  expeditious  and  elegant  methods  of  approxi- 
mation are  furnished  by  the  Calculus.  The  following  is  the  value  of  tc  extended 
10  fifteen  places  of  decimals :    3.141592653589793. 


PROPOSITION  XII. 

374.  TJieoretn, — The  circumference  of  any  circle  is  %7tr,  r 
being  the  radius. 

Dem.— The  circumferences  of  circles  being  to  each  other  as  their  radii,  and 


160  ELEMENTARY  PLANE  GEOMETRY. 

27t    being    the    circumference    of    a    circle    whose    radius    is  1,    we    have 
27t  :  circf.  :  :  1  :  r,  whence  circf.  —  27tr. 

375.  Cor. — The  circumference  of  a  circle   is  ttD,  D  being  the 
diameter,  since  2;rr  =  ;r2r  =  nD. 


ABEA  OF  THE  CIRCLE- 


PROPOSITION  xni. 

37 6 »  TJieorem, — The  area  of  a  circle  ivhose  radius  is  l,is  n, 

Dem.— The  area  of  a  circle  is  |  r  x  circumference  {328).    When  r  =  1,  dr^ 
cumference  =  2it  {372) ;  hence 

area  of  circle  icTiose  radius  isl  ■=  \x2Tt  =  Tt.    q.  e.  d. 


PROPOSITION  xiy. 

577.  Theorem, — Tlie  area  of  any  circle  is  nr^,  r  leing  the 
radius.  "'    - 

Dem. — The  areas  of  circles  being  to  each  other  as  the  squares  of  their  radii, 
and  7t  being  the  area  of  a  circle  whose  radius  is  1,  we  have 

TT  :  aj'ea  of  any  circle  :  :  1"  :  ?*", 

whence  area  of  any  circle  =  nr^.    q.  e.  d. 

378,  Cor. — Tlie  area  of  any  sector  is  such  a  f)art  of  the  area  of 
the  circle  as  the  angle  of  the  sector  is  of  four  right  angles. 

370.  ScH. — As  the  value  of  tt  cannot  be  exactly  expressed  in  numbers,  it 
follows  that  the  area  cannot.  Finding  the  area  of  a  circle  has  long  been 
known  as  the  problem  of  Squaring  the  Circle,  i.  e.,  finding  a  square  equal  in  area 
to  a  circle  of  given  radius.  Doubtless  many  hare-brained  visionaries  or  igno- 
ramuses will  still  continue  the  chase  after  the  phantom,  although  it  has  long 
ago  been  demonstrated  that  the  diameter  of  a  circle  and  its  circumference  are 
incommensurable.  It  is,  however,  an  easy  matter  to  conceive  a  square  of  the 
same  area  as  any  given  circle.  Thus,  let  there  be  a  rectangle  whose  base  is 
equal  to  the  circumference  of  the  circle,  and  whose  altitude  is  half  tlie  radius; 
its  area  is  exactly  equal  to  the  area  of  the  circle.  Now,  let  there  be  a  square 
whose  side  is  a  mean  proportional  between  the  altitude  and  base  of  this  rect- 
angle ;  the  area  of  the  square  is  exactly  equal  to  the  area  of  the  circle. 


APPLICATIONS  OF  THE  DOCTRINE  OF  SIMILARITY.  ICl 


EXERCISES. 

1.  Show  that  if  a  chord  of  a  circle  is  conceived  to  revolve,  varying ^ 
_in  length  as  it  revolves,  so  as  to  keep  its  extremities  in  the  circum- 
ference while  it  constantly  passes  through  a  fixed  point,  the  rect- 
angle of  its  segments  remains  constant. 

2.  The  two  segments  of  a  chord  intersected  hy  another  chord  are 
6  and  4,  and  one  segment  of  the  other  chord  is  3.  What  is  the  other 
segment  of  the  latter  chord  ? 

3.  Show  how  Prop's  L,  IL,  and  III.  may  be  considered  as  differ- 
ent cases  of  one  and  the  same  proposition. 

Sxjg's.— By  stating  Propositions  I.  and  II.  tlius,  The  distances  from  the  inter- 
section  ofthelines  to  their  intersections  with  the  circumference,  what  follows?  In 
Fig.  245,  if  the  secant  AO  becomes  a  tangent,  what  does  OD  become? 

4.  In  a  triangle  whose  sides  are  48,  36,  and  50,  where  do  the  bisec- 
tors of  the  angles  intersect  the  sides  ? 

5.  In  the  last  example  find  the  lengths  of  the  bisectors. 

6.  Keview  the  examples  under  {III9  112,  113,  114),  and 

give  the  reasons. 

7.  In  a  circle  whose  radius  is  20,  what  is  the  length  of  the  arc  of 
a  sector  whose  angle  is  30°  ?    What  is  the  area  of  this  sector  ? 

8.  If  a  circle  whose  radius  is  24  is  divided  into  5  equal  parts  by 
concentric  circumferences,  what  are  the  diameters  of  the  several  cir- 
cles ?    If  the  radius  is  r,  and  number  of  parts  71  ? 

9.  I*rob,—To  divide  a  line  in  extreme  and  mean  ratio  ;  that  is, 
so  that  the  whole  line  shall  he  to  the  greater  segment,  as  the  greater 
segment  is  to  the  less. 

Solution.— Let  it  be  proposed  to  divide  the  line  AB  in  extreme  and  mean 
ratio.  At  one  extremity  of  the  line,  as  B,  erect 
a  perpendicular  equal  to  half  the  line,  that  is, 
make  BO  =  ^  AB.  With  0  as  a  centre,  describe 
a  circumference  passing  through  B.  Draw  AO, 
and  take  AC  equal  to  AD.  Then  is  AB  divided 
in  extreme  and  mean  ratio  at  C,  so  that  AB  : 
AC  : :  AC  :  CB.  To  prove  it,  produce  AO  to  E. 
Now,  AE  :  AB  : :  AB  :  AD  {357\  or  by  inver- 
sion, AB  .  AE  : :  AD  :  AB.  By  division,  AB  : 
AE  -  AB  (=  AE  -  DE)  (or  AD)  (=  AC)  : :  AD  (=  AC)  :  AB  -  AD  (=  AB  -  AC) 
(or  CB).    That  is,  AB  :  AC  : :  AC  :  CB. 

11 


16:^ 


ELEMENTAP.Y   PLANE   GEOMETRY. 


10.  I^rob* — To  inscribe  a  regular  decagon  in  a  circle,  and  hence  a, 
regular  pentagon,  and  regular  j^olygons  of  20,  40,  80,  etc.,  sides. 


Solution. 


{a) 


X 


.:^ 


Fig.  254. 


-Divide  the  radius  iu  extreme  and  mean  ratio,  as  at  (a).  Then  is 
the  greater  segment  ac  the  chord  of 
a  regular  inscribed  decagon,  as 
ABCD,ctc.  To  prove  this,  draw  0 A 
and  OB,  and  taking  OfA  =  ac  =  AB, 
a  side  of  the  polygon,  draw  BM. 
Now,  OA  :  OM  : :  OM  :  MA  by  con- 
struction. As  OM  =  AB^  we  have 
OA  :  AB  ; :  AB  :  MA.  Hence,  con- 
sidering the  antecedents  as  belong- 
ing to  the  triangle  OAB,  and  the 
consequents  to  the  triangle  BAM 
we  observe  that  the  two  sides  about  the  angle  A,  which  is  common  to  botli  tri- 
angles, are  proportional,  hence  the  triangles  are  similar  (342).  Therefore,  ABM 
is  isosceles,  since  OAB  is,  and  angle  BMA  =  A  =  OBA,  and  MB  =  BA  =  OM 
This  makes  0MB  also  isosceles,  and  the  angle  O  =  OBM.  Again  the  exterior 
angle  BMA  —  0  +  OBM  =  20;  hence  A,  which  equals  BMA  =  20.  Hence  also 
OBA,  which  equals  A,  =  20.  Wherefore  0  is  i  the  sum  of  the  angles  of  the 
triangle  OAB,  or  i  of  2  right  angles,  =  tV  of  4  right  angles.  The  arc  AB  is, 
therefore  the  measure  of  -^  of  4  right  angles,  and  is  consequently  -A-  of  the 
circumference. 

To  consti'uct  the  pentagon,  join  the  alternate  angles  of  the  decagon.  To 
construct  the  regular  polygon  of  20  sides,  bisect  the  ai-cs  subtended  by  the  sides 
of  the  decagon,  etc. 

11.  The  projection  of  one  line  upon  another  in  the  same  plane 
is  the  distance  between  the  feet  of  two  perpendiculars  let  fall  from 
the  extremities  of  the  former  upon  the  latter.  Show  that  this  pro- 
jection IS  equal  to  the  square  root  of  the  difference  between  the 
square  of  the  line  and  the  square  of  the  difference  of  the  perpen- 
diculars. 

12.  In  the  triangle  ABCi?  being  a  perpendicular  upon  BA,  prove  that 

7)1  +  n  {=  c)  '.  a  +  1)  '.:  a  —  b'.m  —  n. 

State  the  fact  as  a  proposition.  Give  the  neces- 
sary modification  when  the  perpendicular  falls 
without  the  triangle. 

SuG.  a'  —  7n*  =  &*  —  n*,  whence  a'  —  &'  =  m'  —  n",  etc. 

13.  The  three  sides  of  a  triangle  being  4,  5,  and  6,  find  the  seg- 
ments of  the  last  side,  made  by  a  perpendicular  from  the  opposite 


angle. 


Ans.  3.75,  and  2.25. 


SYN-OPSIS. 


163 


14.  Same  as  above  when  the  sides  are  10, 4,  and  7,  and  the  perpen- 
dicular is  let  fall  from  the  angle  included  by  the  sides  10  and  4. 
Draw  the  figure.     Why  is  one  of  the  segments  negative  ? 

15.  What  is  the  area  of  a  regular  octagon  inscribed  in  a  circle 
whose  radius  is  1?  What  is  its  perimeter?  What  if  the  radius 
is  10? 

16.  What  is  the  side  of  an  equilateral  triangle  inscribed  in  a  circle 
whose  radius  is  1  ? 


SYNOPSIS, 


W 

H 
O 

CO 

» 

«  H 
I— ( 

sg 

«  O 
02  ., 
W    ^ 

H  »-( 

«  23 

W  H 

ft,  O 

O  O 

5 

K 

H 

O 


Importance  of  this  doctrine. 


1 


S  a  ^ 

2  s  z 

H  H  W 

w  o  § 


Prop.  I.  Of  chords. 

Prop.  II.  Of  secants. 

Prop.  III.  Of  secants  and  tangents. 


o 

U  H  J 
W  fc  o 


S  «  to 

<=>  ^  w 
cc  J  s; 


<  o 


!3  w 


Prop.  IV.  How  divide  sides. 
Prop.  V.  Of  exterior  angles. 
Prop.  VI.  Length  of  in  relation  to  other  parts. 
Prop.  VII.  All  intersect  at  one  point. 
l'  Prop.  VIII.  Of  triangles. 


Prop.  IX.  Of  polygons.  < 

Definition  of  rectification. 

Prop.  X.  Perimeters  of 
similar  poly- 
gons. 

Prop.  XI.  Rectification   of  circum- )  c^i 
ference    whose    radius  >• 
isl.  ) 

Prop.  XII.  Circumference  of  any  circle 
=  27rr. 


Cor.  1.  As  squares  of  diagonals. 
C'(9?'.  2.  Regular  polygons. 
Cor.  3.  As  squares  of  apothems. 
Cor.  4:.  Of  circles. 


Cor.  1.  Regular  polygons. 
Cor.  2.  Circumferences  as  radii. 


Signification  and 
importance  of  tc. 


^Cor. 


Also  icD. 


Prop.  XIII.  Whose  radius  is  1 
Prop.  XTV.  Of  any  circle,    j 

Exercises. 


Cor.  Of  sector. 

Sch.  Squaring  the  circle. 


Prob.  To  divide  a  line  in  extreme  and  mean  ratio. 
Prob.  To  inscribe  a  regular  decagon,  etc. 


164  ELEMENTARY   SOLID   GEOMETRY. 


OHAPTEE  II. 

SOLID  GEOMETRY.* 


SECTION  I 

OF  STRAIGHT  LINES  AND  PLANES. 


PERPENDICULAR  AND  OBLIQUE  LINES. 

880*  Solid  Geometry  is  that  department  of  geometry  in 
which  the  forms  (or  figures)  treated  are  not  limited  to  a  single 
plane. 

381,  A  JPlane  (or  a  Fla7ie  Surface)  is  a  surface  such  that  a 
straight  line  joining  two  points  in  it  lies  wholly  in  the  surface. 

III.— The  surface  of  the  blackboard  is  designed  to  be  a  plaDC.  To  ascertain 
■whether  it  is  truly  so,  take  a  ruler  with  a  straight  edge,  and  apply  this  edge  in 
all  directions  upon  it.  If  it  always  coincides,  i.  e.,  touches  throughout  its 
whole  length,  the  sui-face  is  a  plane.  Is  the  surface  of  the  stove-pipe  a  plane  ? 
"Will  a  straight  line  coincide  with  it  in  any  direction?  Will  it  in  every 
direction  ? 


PROPOSITION  I. 

382.  Theorem, — Tliree  points  not  in  the  same  straight  line 
determine  the  position  of  a  i^lane, 

Dem.— Let  A,  B,  and  C  be  three  points  not  in  the  same  straight  line ;  then 
one  plane  can  be  passed  through  them,  and  only  one  ;  i.  e.,  they  determine  the 
position  of  a  plane. 


*  In  some  respects,  perhaps,  "Geometry  of  Space"  is  preferable  to  this  term;  but,  as 
neither  is  free  from  objections,  and  as  this  has  the  advantage  of  simpUcity  and  long  use,  the 
author  prefers  to  retain  it. 


OF  STRAIGHT  LINES  AND   PLANES.  165 

For,  pass  a  straight  line  through  any  two  of  these 
points,  as  A  and   B.    Now,  conceive  any   plane  con-  'C 

taining  these  two  points;  then  will  the  line  passing 
through  them  lie  wholly  in  the  plane  (381).  Con- 
ceive this  plane  to  revolve  on  the  line  as  an  axis  until 
the  point  C  falls  in  the  plane.  Thus  we  have  one 
plane  passed  through  the  three  points.  That  there 
can  be  only  one  is  evident,  since  when  C  tails  in  the 
plane,  if  it  be  revolved  either  way,  C  will  not  be  in  it.  ^^^-  255. 

The  same  may  be  shown  by  first  passing  a  plane  through  B  and  C,  or  A  and  C. 
There  is,  therefore,  only  one  position  of  the  plane  in  which  it  will  contain  the 
third  point,    q.  e.  d. 

383.  Cor.  1. — Through  one  line,  or  two  points,  an  infinite  nuni' 
her  of  planes  can  he  passed, 

384:.  Cor.  2. — Two  intersecting  lines  determine  the  position  of  a 
plane. 

Dem. — For,  the  point  of  intersection  may  be  taken  as  one  of  the  three  points 
requisite  to  determine  the  position  of  a  plane,  and  any  other  two  points,  one  in 
each  of  the  lines,  as  the  other  two  requisite  points.  Now,  the  plane  passing 
through  these  points  contains  the  lines,  for  it  contains  two  points  in  each  line. 

385.  Cor.  S.^Ttvo  parallel  lines  determine  the  position  of  a 
plane. 

Dem.— For,  pass  a  plane  through  one  of  the  parallels,  and  conceive  it 
revolved  until  it  contains  some  point  of  the  second  parallel ;  then  as  the  plane 
cannot  be  revolved  either  way  from  this  position  without  leaving  this  point 
without  it,  it  is  the  only  plane  containing  the  first  parallel  and  this  point  in  the 
second.  But  parallels  lie  in  the  same  plane  {66),  whence  the  plane  of  the  par- 
allels must  contain  the  first  line,  and  the  specified  point  in  the  second.  There- 
fore, the  plane  containing  the  first  line  and  a  point  in  the  second  is  the  plane  of 
the  parallels,  and  is  fixed  in  position. 

386.  Cor.  4. —  The  intersection  of  tioo  pla^ies  is  a  straight  line. 

For  two  planes  cannot  have  even  three  points,  not  in  i7ie  same  straight  line^ 
common,  much  less  an  indefinite  number,  which  would  be  required  if  we  con- 
ceived the  intersection  (that  is,  the  common  points)  to  be  any  other  than  a 
straight  line. 


387.  A  I^erpenclicular  to  a  JPlane  is  a  line  which  is 
perpendicular  to  all  lines  of  the  plane  passing  through  its  foot. 
Conversely,  the  plane  is  perpendicular  to  the  line. 


166 


ELEMENTARY  SOLID  GEOMETBT. 


PROPOSITION  n. 

388.  Tlieorem. — A  line  which  is  perpendicular  to  two  lines  of 
a  plane,  at  their  intersection,  is  perpendicular  to  the  plane, 

Dem.— Let  PD  be  perpendicular  to  AB  and  CF  at  D  ;  then  is  it  perpendicular 
to  MN,  the  plane  of  the  lines  AB  and  CF. 

Let  OQ  be  any  other  line  of  the  plane  MN,  passing 
through  D.  Draw  FB  intersecting  the  three  linea  AB, 
CF,  and  OQ.  Produce  PD  to  P',  making  P'D  =  PD, 
and  draw  PF,  PE,  PB,  P'F,  P'E,  and  P'B.  Then  Ls 
PF  =  P'F,  and  PB  =  P'B,  since  FD  and  BD  are  per- 
pendicular to  PP',  and  PD  =  P'D  {284:).  Hence,  the 
triangles  PFB  and  P'FB  are  equal  {292)\  and,  if  PFB 
be  revolved  upon  FB  till  P  falls  at  P',  PE  will  fall  in 
P'E.  Therefore  OQ  has  E  equally  distant  from  P  and 
p',  and  as  D  is  also  equidistant  from  the  same  points, 
OQ  is  pei-pendicular  to  PD  at  D  {130).  Now,  as  OQ 
is  any  line,  PD  is  perpendicular  to  any  line  of  the 
Fig.  226.  plane  passing  through  its  foot,  and  consequently  per- 

pendicular to  the  plane  {387).    Q-  e.  d. 

389.  Cor. — If  one  of  two  perpendiculars  revolves  alout  the  other 
as  an  axis,  its  path  is  a  plane  perpe?idicular  to  the  axis. 

Thus,  if  AB  revolves  about  PP'as  an  axis,  it  describes  the  plane  MN. 


P 

f. 

M 

iK 

/  K 

'J>^       1 

p^ 

tri 

P' 

f 

PROPOsmox  nii 

300,  Theorem, — At  any  point  in  a  plane  one  perpendicular 
can  he  erected  to  the  plane,  and  only  one. 

Dem. — Let  it  be  required  to  show  that  one  perpen- 
p  ^  dicular,  and  only  one,  can  be  erected  to  the  plane 

MN  at  D.  Through  D  draw  two  lines  of  the  plane, 
as  AB  and  CE,  at  right  angles  to  each  other.  CE 
being  perpendicular  to  AB,  let  a  line  be  conceived  as 
starting  from  the  position  ED  to  revolve  about  AB  as 
an  axis.  It  will  remain  perpendicular  to  AB  {389). 
Conceive  it  to  have  passed  to  P'D.  Now,  as  it  con- 
tinues to  revolve,  P'DC  diminishes  continuously,  and 
at  the  same  rate  as  P'DE  grows  greater;  hence,  in 
one  position  of  the  revolving  line,  and  in  only  one,  as  PD,  PDE  will  equal 
PDC,  and  PD  will  be  perpendicular  to  CE.  Therefore,  PD  is  peri:)endicular  to 
two  lines  of  the  plane,  at  their  intersection,  and  is  the  only  line  that  can  be 
thus  perpendicular,  whence  it  is  perpendicular  to  the  plane  {388),  and  is  the 
only  perpendicular,    q.  e.  d. 


M 

r 

/> 

<rl 

Fig.  227. 


PERPENDICULAR  AND  OBLIQUE  LINES    TO  A  PLANE.  167 

PROPOSITION  IV, 

391*  Theorem, — If  from  any  j^oint  in  a  iierpendicular  to  a 
plane,  oUique  lines  he  drawn  to  the  plane,  those  ichich pierce  the 
plan6  at  equal  distances  from  the  foot  of  the  perpendicular  are  equal; 
and  of  those  ivhicli  pierce  the  plane  at  unequal  distances  from  the 
foot  of  the  perpendicular,  those  lohich  pierce  at  the  greater  distances 
are  the  greater. 

Dem. — Let  PD  be  a  perpendicular  to  the  plane 
MN,  and  PE,  PE',  PE",  and  PE'",  be  oblique  lines 
piercing  the  plane  at  equal  distances  ED,  ED,  E"D, 
and  E'"D,  from  the  foot  of  the  perpendicular;  then 
PE  =  PE'  =  PE"  =  PE'".  For  each  of  the  tri- 
angles PDE,  PDE',  etc.,  has  two  sides  and  the  in- 
cluded angle  equal  to  the  corresponding  parts  in 
the  other. 

Again,  let  FD  be  longer  than  E'D.  Then  is 
PF>  PE'.    For,  take  ED  =  E'D  ;   then  PE  =  PE',  Fia.  258. 

by  the  preceding  part  of  the  demonstration.     But 
PF  >  PE  by  {139).    Hence,  PF  >  PE'.    q.  e.  d. 

392,  The  Inclination  of  a  line  to  a  plane  is  measured  by 
the  angle  which  the  line  makes  with  a  line  of  the  plane  passing 
through  the  point  in  which  the  line  pierces  the  plane  and  the  foot 
of  a  perpendicular  to  the  plane  from  any  point  in  the  line. 

Thus  PFD  is  the  inclination  of  PF  to  the  plane  MN. 

393,  Cor.  1. — The  angles  luhich  oUiquc  lines  draion  from  a  com- 
mon point  in  a  perpendicular  to  a  plane,  and  ptiercing  the  plane  at 
equal  distances  from  the  foot  of  the  perpoidiciilar,  make  with  the 
perpendicular,  are  equal ;  and  the  inclinations  of  such  lines  to  the 
plane  are  equal. 

Thus  the  equality  of  the  triangles,  as  shown  in  the  demonstration,  shows  that 
EPD  =  E'PD  =  E"PD  =  E"'PD,  and  PED  =  PE'D  =  PE"D  =  PE"'D. 

394,  Cor.  2. — Conversely,  If  the  angles  which  oilique  lines 
draion  from  a  point  in  a  perjjendicular  to  a  plane,  make  toith  the 
perpendicular,  are  equal,  the  lines  are  equal,  and  p>ierce  the  plane  at 
equal  distances  from  the  foot  of  the  perpendicular. 

DEM.-Thus,  in  the  figure,  let  DPE'  =  DPE";  then  PE'  =  PE"  and  DE'  = 
DE".  For,  revolve  DE'P  about  PD  ;  DE'  will  continue  in  the  plane  MN,  and 
when  angle  DPE'  coincides  with  its  equal  DPE",  PE'  coincides  with  PE",  and 
DE'  with  DE '. 


168  ELEMENTABY  PLANE  GEOMETRY. 

39o,  Cor.  3. — Also,  conversely,  Equal  ollique  lines  from  the 
same  point  in  the  perpendicular,  pierce  the  plane  at  equal  distances 
from  the  foot  of  thep)erpendicular, 

Dem.— Let  PE'  =  PE" ;  then  is  DE'  =  DE".  For,  let  PDE'  revolve  npon  PD 
until  ED  foils  in  E"D  ;  then,  if  DE'  were  less  thaa  DE",  PE'  would  be  less  than 
PE"  ;  and,  if  DE'  were  greater  than  DE",  PE'  would  be  greater  than  PE".  But 
both  of  these  conclusions  are  contrary  to  the  hypothesis.  Hence,  as  DE'  can 
neither  be  less  nor  greater  than  DE  ",  it  must  equal  it.  This  corollary  follows 
also  from  {297). 

396,  Cor.  4. — The  perpendicular  is  the  shortest  line  that  can  le 
drawn  to  a  plane  from  a  point  without,  and  measures  the  distance  of 
the  p)oint  from  the  p lane. 


PROPOSlllOX  T. 

397 •  TJieoreui, — From  a  point  ivithout  a  plane  one  perpendic- 
ular can  he  drawn  to  the  plane,  and  only  one, 

Dem. — Let  it  be  required  to  show  that  one  perpen- 
dicular can  be  drawn  from  P  to  the  plane  MN,  and 
only  one.  Take  AB,  any  line  of  the  plane,  and  con- 
ceive PD'  perpendicular  to  it.  Through  D'  draw  EF, 
a  line  of  the  plane,  perpendicular  to  AB,  Now,  if 
PDE  =  PD'F,  they  are  both  right  angles,  and  PD'  is 
perpendicular  to  two  lines  of  the  plane  passing  through 
its  foot,  and  hence  perpendicular  to  the  plane  (388). 
If,  however,  PD'E  does  not  equal  PD'F,  in  the  first  in- 
stance, but  PDF  <  PDE,  conceive  the  line  AB  to 
move  along  the  plane,  continuing  parallel  to  its 
primitive  position,  so  as  to  cause  D'  to  move  towards  F,  thus  diminishing  PDE 
and  increasing  PD'F.  At  the  same  time  observe  that,  if  necessary  in  order  to 
keep  PD  A  =  PD'B*,  EF  can  move  along  the  plane  parallel  to  its  first  position. 
Now,  as  PD  F  increases,  passing  through  all  successive  values,  and  PD'E  dimin- 
ishes in  the  same  way,  there  will  be  some  position  of  PD',  as  PD,  in  which 
PDF  =  PDE,  and  as  by  hypothesis  PDA'  remains  =  PDB',  PD  becomes  perpen- 
dicular to  two  lines  passing  through  its  foot,  and  hence  perpendicular  to  the 
plane. 

That  there  can  be  only  one  perpendicular  is  evident,  since,  if  there  were  two, 
as  PD'  and  PD,  there  would  be  two  right  augles  in  the  triangle  PD  D. 

*  According  to  the  conception  here  used  it  would  not  be  necessary. 


PARALLEL     LINES     AND     PLANES.  •  169 

S98,  Cor. — Tlirough  a  given  point  in  a  line  one  plane  can  be 
passed  perpendicular  to  the  line,  and  only  one. 

Dem. — Let  D  be  the  point  in  the  line  PD.  Pass  two  lines  through  D,  as  EF, 
and  A''B',  each  perpendicular  to  PD  ;  the  plane  of  these  lines  is  perpendicular  to 
PD.  Moreover,  the  required  plane  must  contain  both  these  lines,  for  if  it  passed 
through  D  and  did  not  contain  DF,  there  would  be  one  line  of  the  plane,  at 
least,  which  would  pass  through  D  and  not  be  perpendicular  to  PD,  which  is 
impossible.  Hence,  there  can  be  no  other  plane  than  the  plane  of  the  two  per- 
pendiculars EF  and  A'B'  which  shall  be  perpendicular  to  PD,  through  D. 


PROPOSITION  TI. 

399.  Theorem, — If  from  the  foot  of  a  perpendicular  to  a  plane 
a  line  he  drwwn  at  right  angles  to  any  line  of  the  pilcLne,  and  this 
intersection  he  joined  with  any  point  in  the 
•perpendicular,  the  last  line  is  perpendicular  to 
the  line  of  the  plane* 

Dem. — From  the  foot  of  the  perpendicular  PD  let 
DE  be  drawn  perpendicular  to  AB,  any  line  of  the 
plane  MN,  and  E  joined  with  0,  any  point  of  the  per- 
pendicular ;  then  is  OE  perpendicular  to  AB. 

Take  EF  =  EC,  and  draw  CD,  FD,  CO,  and  FO. 
Now,  CD  =  DF  (?)*  whence  CO  =  FO  (?),  and  OE  has 
O  equally  distant  from  F  and  C,  and  also  E.     There-  Fig.  2G0. 

fore,  OE  is  perpendicular  to  AB  (?).    q.  e.  d. 

4:00.  Cor. — The  line  DE  measures  the  shortest  distance  between 
PD  a7id  AB. 

For,  if  from  any  other  point  in  AB,  as  C,  a  line  be  drawn  to  D,  it  is  longer 
than  DE(?) ;  and  if  drawn  from  C  to  a,  any  other  point  in  PD  than  D,  Ca  is 
longer  than  CD  (?),  and  consequently  longer  than  DE  (?). 


PARALLEL  LINES  AND  PLANES. 


401,  A  Line  is  Parallel  to  a  plane  when  the  two  will 
not  meet,  how  far  soever  they  be  prodnced.  The  plane  is  also  said 
to  be  parallel  to  the  line. 

*  Hereafter  the  reason  will  be  often  left  out,  and  the  mark  (?)  will  be  used  to  indicate  that 
the  student  is  to  supply  it. 


170 


ELEMENTARY  SOLID  GEOMETRY. 


;^ 


@ 


PROPOSITION  Til. 


^ 


402.  Hieore^n. — One  of  tim  parallel  lines  is  parallel  to  every 
plane  containing  the  other. 


Fig.  261. 


Dem. — AB  being  parallel  to  CD  is  parallel  to 
1  any  plane  MN  containing  CD. 

/  Since  AB  and  CD  are  in  the  same  plane  (?)^ 

and  as  the  intersection  of  their  plane  with  MN  is 
[sj  CD  (?),  if  AB  meets  the  plane  MN,  it  must  meet 

it  in  CD,  or  CD  produced.     But  this  is  impossi- 
ble (?).     Whence  AB  is  parallel  to  MN.    q.  e.  d. 

403.  Cor.  l.  Tlirough  any  given  line  a  plane  may  be  passed 
parallel  to  any  other  given  line  not  in  the  plane  of  tlie  first. 

For,  through  any  point  of  the  line  through  which  the  plane  is  to  pass,  con- 
ceive a  line  parallel  to  the  second  given  line.  The  i)lane  of  the  two  intersecting 
lines  is  parallel  to  the  second  given  line  (?). 

404,  Cor.  2. — Tlirough  any  point  in  space  a  plane  may  Repassed 
parallel  to  any  two  lines  in  space. 

For,  through  the  given  point,  conceive  two  lines  parallel  to  the  given  lines ; 
then  is  the  plane  of  these  intersecting  lines  parallel  to  the  two  given  lines  (?). 


PROPOSITION  Tm* 

405 •  Theoreifn. — If  07ie  of  two  parallels  is  perpendicular  to  a 
plane,  the  other  is  perpendicular  also. 

Dem.— Let  AB  be  parallel  to  CD,  and  perpendicular  to  the  plane  MN  ;  then 
is  CD  perpendicular  to  MN. 

For  drawing  BD  in  the  plane  MN,  it  is  perpendicular 
to  AB  (?),  and  consequently  to  CD  (?).  Through  D  draw 
EF  in  the  plane  and  perpendicular  to  BD,  and  join  D 
with  any  point  in  AB,  as  A;  then  is  EF  perpendicular 

to  AD  (?).  Now,  EF  being  perpendicular  to  two  lines,  AD 

/    and  BD  of  the  plane  ABDC,  is  perpendicular  to  the  plane, 
/    "  y^    I      and  hence  to  any  line  of  the  plane  passing  through  D, 

^ ~ — 4vl       as  CD.    Therefore  CD  is  perpendicular  to  BD  and  EF, 

■   Fig  262.  and  consequently  to  the  plane  MN  (?).    q.  e.  d. 

40G.  Cor.  1. — Two  lines  which  are  perpendicular  to  the  same 
plane  are  parallel. 

Thus,  AB  and  CD  being  perpendicular  to  the  plane  MN,  if  AB  is  not  parallel 
to  CD,  draw  a  line  through  B  which  shall  be.  By  the  proposition  this  Ime  is 
perpendicular  to  MN,  and  hence  must  coincide  with  AB  {390). 


M 


PARALLEL  LINES  AND   PLANES. 


171 


4:07*  Cor.  2. — Two  lines  parallel  to  a  third 
not  in  their  oiun  plane  are  parallel  to  each  other. 

Dem. — If  AB  and  CD  are  parallel  to  EF,  they  are  par- 
allel to  each  other.  Let  MN  be  a  plane  passing  through 
EF  at  F',  and  to  which  EF  is  perpendicular;  then  areAB 
and  CD  respectively  perpendicular  to  MN  (?),  and  hence 
parallel  (?).    Q.  e.  d. 


;: — TN 


B  ^ 

Fig.  263. 


408,  Parallel  Planes  are  such  as  do  not  meet  when  indefi- 
nitely produced. 


PROPOSITION  IX. 
409,  Theorem. — Two  planes  perpendicular  to  the  same  line  are 
parallel  to  each  other, 

Dem.— For,  if  they  could  meet  in  some  point,  as  0,  conceive  two  lines  drawn 
from  O,  one  in  each  plane,  to  the  points  where  the  perpendicular  pierces  the 
planes.  We  should  then  have  two  lines  from  the  same  point,  perpendicular  to 
the  same  line(?),  which  is  impossible.  Hence,  as  the  planes  cannot  meet,  they 
are  parallel.    Q.  e.  d. 


PROPOSITION  X. 

4^0.  Theorem* — If  a  plane  intersect  two 
parallel  planes^  the  lines  of  intersection  are 
parallel, 

Dem. — Let  AD  intersect  the  parallel  planes  MN  and 
PQ  in  AB  and  CD  ;  then  is  AB  parallel  to  CD.  For,  if 
AB  and  CD  could  meet,  the  planes  MN  and  PQ  would 
meet,  as  every  point  in  AB  is  in  MN,  and  every  point 
in  CD  in  PQ.  Hence,  AB  and  CD  lie  in  the  same 
plane,  and  do  not  meet  how  far  soever  they  be  pro- 
duced ;  they  are  therefore  parallel.    Q.  e,  d. 

411,  Cor. — Parallel  lines  intercepted  hetioeen  parallel  planes  are 
equal. 

Thus  AC  "BD  if  they  are  parallel.  For,  the  intersections  AB  and  CD,  of 
the  plane  of  these  parallels,  are  parallel  (?),  and  the  figure  A B DC  is  a  paral- 
lelogram ;  whence  AC  =  BD  (?). 


Fig.  264. 


PROPOSITION  XI. 
412,  Theorem, — A  line  which  is  perpendicular  to  one  of  two 
par aUcl  planes,  is  perpendicular  to  the  other  also. 


172 


ELEMENTARY  SOLID  GEOMETRY. 


M. 


'N 


B 


Dem. — Let  AB  be  perpendicular  to  MN  ;  then  is  it  perpendicular  to  PQ,  paral- 
lel to  MN.      For,  pass  two  planes  through  AB,  and  let 
£-.  Y     CA,  DB,  and  EA,  FB,  be  their  intersections  with  the 

planes  MN  and  PQ.  Now  CA  and  EA  are  perpendicu- 
lar to  AB  (?) ;  hence  DB  and  FB  being  parallel  to  CA 
and  EA  (?)  are  pei-pendicular  to  AB  (?).  Therefore  AB 
is  perpendicular  to  two  lines  of  the  plane  PQ,  and 
consequently  to  the  plane  (?).    q.  e.  d. 

4:13.  Cor.  1. — Tlirough  any  'point  out  of  a 

'^  °  "" Q  plane,  one  plane  can  he  passed  parallel  to  the 

Fig.  265.  given  plane,  and  only  one. 

Dem. — To  pass  a  plane  through  B  parallel  to  MN,  draw  the  perpendicular 
BA  from  B  upon  MN.  Draw  any  two  Imes  in  the  plane  MN,  through  A,  as  CA 
and  EA.  Through  B  draw  DB  and  FB  parallel  to  CA  and  EA;  then  is  PQ,  the 
plane  of  these  lines,  perpendicular  to  AB,  and  hence  parallel  to  MN.  As  the 
plane  parallel  to  MN  must  contain  FB  and  DB,  and  as  but  one  plane  can  be 
passed  through  these  lines,  there  can  be  only  one  plane  through  B  parallel  to  MN. 

4:14,  Tfie  JDistance  hetween  two  parallel  planes  at  any  point 
is  measured  iy  the  perpendicular, 

415 •  Cor.  2. — Parallel  planes    are    every- 
where equally  distant  from  each  other. 


De3I. — Let  A  and  B  be  any  two  points  in  the  plane 
MN,  and  AC  and  BD  tlie  perpendiculars  from  these 
points,  let  fall  on  the  parallel  plane  PQ  ;  then  are  they 
perpendicular  to  MN  by  the  proposition,  and  since  the 
figure  ABCD  is  a  parallelogram  (?)  [a  rectangle,  also  (?)], 
Fig.  266.  AC  =  BD. 


PROPOSITION  xn. 

416.  TJieorem. — Two  angles  lying  in  different  planes,  hut  hav- 
ing their  sides  parallel  and  extending  in  the  same 
direction,  are  equal,  and  their  planes  are  parallel 

Dem.— Let  A  and  A'  lie  in  the  different  planes  MN  and 
PQ,  and  have  AB  parallel  to  A'B',  and  AC  to  A'C ;  then 
A  =  A',  and  M  N  and  PQ  are  parallel. 

For,  take  AD  =  A'D',  and  AE  =  A'E',  and  draw  AA', 
DD',  EE',  ED,  and  E'D'.  Now,  AD  being  equal  and  par- 
allel to  A'D',  AA'  =  DD'  (?).  For  like  reason  AA'  =  EE', 
therefore  EE'  —  DD'.  Again,  since  EE'  and  DD'  are 
respectively  parallel  to  AA',  they  are  parallel  to  each 
other  (?),  whence  EDD'E'  is  a  parallelogram  (?),  and 
ED  =  E'D'.     Hence  the  triangles  ADE  and  A'D'E'  are 


Fig.  267. 


OF  STRAIGHT  LINES  AND  PLANES. 


173 


mutually  equilateral,  and  A,  opposite  ED,  is  equal  to  A',  opposite  E'D'  equal 
to  ED. 

Again,  the  plane  of  the  angle  BAC,  MN,  is  parallel  to  PQ,  the  plane  of  B'A'C. 
For,  let  a  plane  be  passed  through  AC  and  revolved  until  it  is  parallel  to  PQ. 
It  must  cut  DD',  which  is  parallel  to  AA',  and  EE',  so  that  DD'  shall  equal  AA' 
and  EE'  (?);  hence  it  must  pass  through  D. 

4zl7,  Cor.  1. — Ifttoo  intersecting  planes  lecut  ly  parallel  planes, 
the  angles  formed  ly  the  i^itersections  are  equal, 

Th«s,  AB'  and  AE'  being  cut  by  the  parallel  planes  MN  and  PQ,  AD  is  parallel  to 
A'D'  (?),  and  lies  in  the  same  dh'ection,  and  AE  to  AE'.    Hence   BAC=  B'A'C  (?). 

418,  Cor.  2. — If  the  corresponding  exiremities  of  three  eqical 
parallel  lines  not  in  the  saine  plane,  be  joined,  the  triangles  formed 
are  equal,  and  their  planes  parallel. 

Thus,  if  AA'  =  DD'  =  EE',  the  sides  of  the  triangle  AED  are  equal  to  the 
sides  of  A'E'D',  since  the  figures  AD',  DE',  and  EA'  are  parallelograms  (?),  and 
the  corollary  comes  under  the  proposition  (?). 


PROPOSITION  XIII. 

410,  Theorem, — The  corresponding  seg- 
ments of  lines  cut  by  parallel  planes  are  propor- 
tional. 

Dem.— LetAB.CD  and  EF  be  cut  by  the  parallel  planes 
MN,  PQ,  RS,  and  TU  ;  then  Aa  :  Cd  : :  a6  :  «/  : :  5B  :  /D, 
and  Aa  :  Ei  : :  ah  :ik  ::  bB  :  kF,  and  Ce  :  Ei  : :  ef  :  ik  : : 
/D  :kF. 

For,  join  the  extremities  A  and  D,  and  E  and  D,  and 
conceive  the  intersections  of  the  plane  of  AB  and  AD 
with  the  parallel  planes  to  be  BD,  bd,  and  ac.  These 
lines  are  parallel  (?),  and  Aa  :  fiiC  ::  ab  :  cd  ::  bB  :dD  (?). 
For  a  similar  reason,  Ce  :  Ac  ::  ef  :  cd  ::  fD  :  dD  (?). 
Whence,  the  consequents  of  the  proportions  being  the 
same,  the  antecedents  give  Aa  :  Ce  :  :  ab  :  ef  ::  bB  :  /D. 
In  like  manner  we  can  show  that  Ce  :  Ei  ::  ef :  ik  :  :/D  : 
kF.  [Let  the  student  give  the  details.]  From  these 
proportions  we  have  Aa  :  Ei  ;:  ab  :  ik  : :  bB  :  kF  (?). 
Q.  E.  D. 


lt>/. 

pf 

;/     '^ 

I 

^  1  1 

1^ 

R 

!   I      Q 

f-^ 

UL 

T 

/  ' 

M 

' '- 1 

Fig.  268. 


EXERCISES. 

1.  Designate  any  three  points  in  the  room,  as  one  corner  of  the 
desk,  a  point  on  the  stove,  and  some  point  in  the  ceiling,  and  show 
how  you  can  conceive  the  plane  of  these  points. 


ft 

174:  ELEMENTARY   SOLID   GEOMETRY. 

2.  Show  the  position  of  two  lines  which  will  not  meet,  and  yet  are 
not  parallel. 

3.  Conceiye  two  lines,  one  line  in  the  ceiling  and  one  in  the  floor, 
which  shall  not  be  parallel  to  each  other.  What  is  the  shortest  dis- 
tance between  these  lines  ? 

4.  The  ceiling  of  my  room  is  10  feet  above  the  floor.  I  have  a  12 
foot  pole,  by  the  aid  of  which  I  wish  to  determine  a  point  in  the  floor 
directly  under  a  certain  point  in  the  ceiling.    How  can  I  do  it  ? 

SuG.— Consult  Prop.  IV. 

5.  Upon  what  principle  in  this  section  is  it  that  a  stool  with  three 
legs  always  stands  firm  on  a  level  floor,  when  one  with  four  may 
not? 

6.  By  the  use  of  two  carpenter's  squares  you  can  determine  a  per- 
pendicular to  a  plane.    How  is  it  done  ? 

7.  If  yon  wish  to  test  the  perpendicularity  of  a  stud  to  a  level  floor, 
on  how  many  sides  of  it  is  it  necessary  to  measure  the  angle  which  it 
makes  with  the  floor  ?  By  applying  the  right  angle  of  the  carpen- 
ter's square  on  any  two  sides  of  the  stud,  to  test  the  angle  which  ^t 
makes  with  the  floor,  can  you  determine  whether  it  is  perpendicular 
or  not  ? 

8.  We  see  in  straight  lines.  If  a  line*  be  placed  between  our  eye 
and  a  surface,  it  covers  a  certain  space  on  the  surface ;  this  figure  or 
space  is  said  to  be  the  projection  of  the  line  on  that  surface.  Upon 
what  principles  in  this  section  is  it  that  the  projections  of  straight 
lines  are  straight  ?  Why  is  it  that  the  projections  of  parallels  which 
are  parallel  to  the  plane  upon  which  we  see  them  projected,  are 

■parallel,  while  parallel  lines  which  are  inclined  to  this  plane  are  pro- 
jected in  oblique  lines  ? 

9.  If  a  line  is  drawn  at  an  inclination  of  23°  to  a  plane,  what  is 
the  greatest  angle  which  any  line  of  the  plane,  drawn  through  the 
point  where  the  inclined  line  pierces  the  plane,  makes  with  the  line  ? 
Can  you  conceive  a  line  of  the  plane  which  makes  an  angle  of  50° 
with  the  inclined  line?     Of  80°  ?     Of  15°  ?     Of  170°  ? 

Sereafter^the  student  should  make  the  sxjnopses. 


*  The  term  line  is  here  ased  in  its  colloquial  sense,  and  refers  to  a  material  representation, 
as  a  cord,  the  edge  of  a  board,  etc. 


OF  SOLID  ANGLES. 


175 


SECTION  II. 

OF    SOLID    ANGLES. 


4:20  •  A  SolictJLngle^  is  the  opening  between  :two  or  more^ 
planes,  each  of  which  intersects  alFlhe  others.    The  lines  of  intei'.-^ 
section  are  called  .£^£65,  and  the  planes,  or  the  portion  of  the  planes 
between  the  edges,  where  there  are  more  than  two,  are  called  Faces.^ 

421.  A  Dieclral  Angle,  or  simply  a  Diedral,  is  the  opening 
between  hvo  intersecting  planes. 

422.  A  Polyedral  Angle,  called  also  simply  a  Polyedral,  is 
the  opening  between  three  or  more  planes  which  intersect  so  as  to 
have  one  common  point,  and  only  one.  In  the  case  of  three  inter- 
secting planes  the  angle  is  called  a  Triedral.  The  point  common  to 
all  the  planes  is  called  the  Vertex.  The  plane  angles  enclosing  a 
polyedral  are  the  Facial  angles. 

423.  A  Diedral  {Angle)  is  measured  by  the  plane  angle  included 
by  lines  drawn  in  its  faces  from  any  point  in  the  edge,  and  perpen- 
dicular thereto.  A  diedral  angle  is  called  right,  acute,  or  obtuse, 
according  as  its  measure  is  right,  acute,  or  obtuse.  Of  course  the 
magnitude  of  a  solid  angle  is  independent  of  the  distances  to  which 
the  edges  may  chance  to  be  produced. 

Ill's.— The  opening  between  the  two  planes  CABF  and  DABE  is  a  Diedral 
(angle),  AB  is  the 
Edge,  and  CABF 
and  DABE  are  the 
Faces.  Let  MO 
lie  in  the  plane 
AF,  perpendicular 
to  the  edge;  and 
NO  in  AE,  and  also 
perpendicular  to 
the  edge ;  then  the 
plane  angle  MON 
is  the  measure  of 
the  diedral.     A  diedral  may  be  read  by  the  letters  on  the  edge,  -when  there 


176 


ELEMENTARY  SOLID  GEOMETEY. 


would  be  no  ambiguity,  or  otherwise  by  these  letters  and  one  in  each  face"; 
thus,  tht^  diedral  in  (a)  may  be  designated  as  AB,  or  as  C-AB-D. 

In  (6)  we  have  a  Ti-iedral  (angleX,  The  edges  are  SA,  SB,  and  SC;  HolQ  faces 
ASB,  BSC,  and  ASC:  the  facial  angles  are  ASB,  ASC,  and  BSC ;  and  S  is  the 
vertex.  Such  an  angle,  and  any  polyedral  (angle),  may  be  read  by  naming  the 
angle  at  the  vertex,  when  there  would  be  no  ambiguity,  or  otherwise  by  naming 
the  letter  at  the  vertex,  and  then  one  in  each  edge ;  thus  S-ABCDE  designates 
the  polyedral  (o).    The  openmg  between  the  planes  is  the  angle,  in  each  case. 


OF  DIEBBALS. 

4:24:»  A  Diedral  may  be  considered  as  generated  by  the  revolution 
of  a  plane  about  a  line  of  the  plane,  and  hence  we  may  see  the  pro- 
priety of  measuring  it  by  the    angle    included  by 
two  lines  in  its  faces  perpendicular  to  its  edge,  as 
stated  in  the  preceding  article. 

III. — Let  A3  be  a  line  of  the  plane  GB.  Conceive  gB 
perpendicular  to  AB.  Now,  let  the  plane  revolve  upon  AB 
as  an  axis,  whence  gB  describes  a  circle  (?) ;  and  at  any  posi* 
tion  of  the  revolving  plane,  as/BAF,  since  fBg  measures  the 
amount  of  revolution,  it  may  be  taken  as  the  measure  of  the 
diedral  f-Bk-g.  When^-B  has  made  i  of  a  revolution,  the 
plane  will  have  made  i  of  a  revolution,  and  the  diedral  will 
be  right. 

425,  Coit. — Opposite  diedrals  are  equal. 

Thus,  if  C-AB-D  is  measured  by  MON,  c-AB-d  is  measured 
by  the  equal  angle  nOm. 


PROPOSITION  1. 

4:26,  TJieorem. — Any  line  in  one  face  of  a  rigid  diedral,  per- 
pendicular to  its  edge,  is  perpendicular  to  the  other 
face, 

Dem.— In  the  face  CB  of  the  right  diedral  C-AB-D,  let  MO 
be  perpendicular  to  the  edge  AB  ;  then  is  it  perpendicular  to 
the  face  DB.  For,  draw  ON  in  the  face  DB,  and  perpendicu- 
lar to  AB.  Now,  since  the  diedral  is  right,  and  MON  meas- 
ures its  angle,  MON  is  a  right  angle;  whence  MO  is  perpen- 
dicular to  two  lines  of  the  plane  DB,  and  consequently  per- 
Fio.  272.  pendicular  to  the  plane,     q.  e.  d.  ' 


OF  DIEDRALS. 


177 


427 •  Cor. — Conversely,  If  one  plane  contain  a  line  which  is 
perpendicular  to  another  plane,  the  diedral  is  right. 

Thus,  if  MO  is  perpendicular  to  the  plane  DB,  C-AB^D  is  a  right  diedral. 
For  MO. is  perpendicular  to  every  line  of  DB  passing  tlirough  its  foot  (?);  and 
hence  is  perpendicular  to  ON,  drawn  at  right  angles  to  AB.  Whence  C-AB-D  is 
a  right  diedral,  for  it  is  measured  by  a  right  plane  angle. 


M 


PROPOSITION  II. 

428.  Theo7*ein, — If  tivo  planes  are  perpendicular  to  a  third, 
their  intersection  is  peipendicular  to  the  tidrd p>lcine. 

Dem. — If  CD  and  EF  are  perpendicular  to  the  plane 
MN,  then  is  AB  perpendicular  to  MN.  For,  EF  being 
perpendicular  to  MN,  D-FG-E  is  a  right  diedral,  and  a 
line  in  EF  and  perpendicular  to  FC  at  B  is  perpendicular 
to  MN  ;  also  a  line  in  the  plane  CD,  and  perpendicular  to 
DH  at  B,  is  perpendicular  to  MN  (?).  Hence,  as  there 
can  be  one  and  only  one  perpendicular  to  MN  at  B,  and 
as  this  perpendicular  is  in  both  planes,  CD  and  EF,  it 

Q.  E.  D. 


N 


Fig.  273. 

is  their  intersection. 


PROPOSITION  III. 

420.  TJieorem, — If  from  any  point  perpendiculars  he  drawn 
to  the  faces  of  a  diedral  angle,  their  included  angle  will  be  the  supple- 
ment of  the  angle  lohich  measures  the  diedral,  or  equal  to  it. 

Dem. — Let  BD  and  AD  be  any  two  planes  including  the  ^ 

diedral  A-SD-B,  then  will  two  lines  drawn  from  any  point, 
perpendicular  to  these  planes,  include  an  angle  which  is  the 
supplement  of  the  measure  of  the  diedral,  or  equal  to  it. 

If  the  point  from  which  the  lines  are  drawn  is  not  in 
the  edge  SD,  we  may  conceive  two  lines  drawn  through 
any  point,  as  S,  in  this  edge,  which  shall  be  parallel  to  the 
two  proposed,  and  hence  include  an  equal  angle,  and 
have  their  plane  parallel  to  the  plane  of  the  proposed 
angle  {416).  Let  the  latter  lines  be  SO  and  SP.  We  are 
to  show  that  OSP  is  supplemental  to  the  measure  of  A-SD-B. 
A  plane  passed  through  S,  perpendicular  to  the  edge  SD, 
will  contain  the  lines  SO  and  SP  {388) ;  and  its  intersec- 
tions with  the  faces,  as  SB  and  SA,  will  form  an  angle 
(ASB)  which  is  the  measure  of  the  diedral  {423).  Now, 
PSA  =  a  right  angle  (?),  and  OSB  =  a  right  angle  (?).  Hence, 
PSO  and  ASB  are  either  equal  or  supplemental  {283). 

Q.  E.  D. 


178 


ELEMENTARY   SOLID   GEOMETRY. 


430.    Triedrals 


^^^ — 

'--._^^,^^ 

B 

■!'■ 

Fig.  275. 


OF   TRIEDRALS. 

are  Rectangular,  Biredangular,  or  Trirectan- 
gnlar,  according  as  they  have  one,  two,  or  three, 
right  diedral  angles. 

Ill's. — The  corner  of  a  cube  is  a  Trirectangular 
triedral,  as  S-ADC.  Conceive  the  upper  portion  ol" 
the  cube  removed  by  the  plane  ASEF;  then  the  angle 
at  S,  i.  e.,  S-AEC  is  a  Biredangular  triedi-al,  A-SC-E 
and  A-SE-C  being  right  diedrals. 


431,  An  Isosceles  Triedral  is  one  that  has  two  of  its  facial 
angles  equal.  An  Bquilateral  Triedral  is  one  that  has  all 
three  of  its  facial  angles  equal. 

432,  Two  Sf/minetrical  Triedrals  are  suoh  as  have  the 
facial  angles  of  the  one  equal  to  the  facial  angles  of  the  other,  each 
to  each;   but  in  which  the  equal   facial   angles  are   not  similarly 

situated,  and  hence  the  triedrals  are  not  necessarily 
capable  of  superposition. 

Ill's. — Let  the  edges  of  the  triedral  S-ABC,  be  pro- 
duced beyond  the  vertex,  forming  a  second  triedral  S-abc ; 
then  are  the  two  triedrals  symmetrical,  i.  e.,  the  faces 
are  equal  plane  angles,  but  disposed  in  a  different  order. 
Thus,  ASB  =  aSb,  ASC  =  aSc,  and  BSC  =  bSc ;  but  the 
triedrals  cannot  be  made  to  coincide.  To  show  this 
fact,  conceive  the  upper  triedral  detached,  and  the  face 
aSc  placed  in  its  equal  face  ASC,  Sa  in  SA,  and  Sc  in  SC. 
Now,  the  edge  Sb,  instead  of  falling  in  SB  will  fall  on  the 
left  of  the  plane  ASC. 

Sj'mmetrical  solids  are  of  frequent  occurrence:  the  two 
hands  form  an  illustration ;  for,  though  the  parts  may  be 
exactly  alike,  the  hands  cannot  be  placed  so  that  their 
like  parts  will  be  similarly  situated;  in  short,  the  left 
glove  will  not  fit  the  right  hand. 

433,  Two  triedrals  are  SiiJU^lementary  when  the  edges  of 
one  are  perpendicular  to  the  faces  of  the  other.  (See  438a,) 


Fig.  '2\\i. 


PROPOSITION  IV. 

434,  Tlieoreni, — The  sum  of  any  tvjo  facial  a^igles  of  a  trie' 
dral  is  greater  than  the  third. 


OF  TRIEDRALS. 


179 


Dem. — This  proposition  needs  demonstration  only  in 
case  of  the  sum  of  the  two  smaller  facial  angles  as  com- 
pared with  the  greatest  (?).  Let  ASB  and  BSC  each  be  less 
than  ASC ;  then  is  ASB  +  BSC  >  ASC.  For,  make  the  an- 
gle ASb''  =  ASB,  and  Sb'  =  S6,  and  pass  a  plane  through  b 
and  &',  cutting  SA  and  SC  in  a  and  c.  The  two  triangles  aSb 
and  aSb'  are  equal  (?),  whence  ab'  =  ab.  Now,  ab  +  be  >  ac 
(?),  and  subtracting  ab  from  the  first  member,  and  its  equal 
(d>'  from  the  second,  we  have  be  >  b'c.  Whence  the  two  tri- 
angles bSc  and  b'Sc  have  two  sides  equal,  but  the  third  side 
be  >  than  the  third  side  b'e,  and  consequently  angle  bSc  > 
b'Sc.  Adding  ASB  to  the  former,  and  its  equal  AS5'  to  the 
latter,  we  have  ASB  +  BSC  >  ASC.    q.  e.  d. 


Fig.  a77. 


43 S,  Cor. — The  difference  hettoeen  any  tioo  facial  angles  of  a 
triedral  is  less  than  the  third  facial  angle  (?). 


PROPOSITION  V. 

436,  Theorem, — The  sum  of  the  facial  angles  of  a  triedral 
mag  he  anything  hePween  0  and  four  right  angles. 

Dem.— Let  ASB,  BSC,  and  ASC   be  the  facial  angles 
enclosing  a  triedral ;  then,  as  each  must  have  some  value,  ,.'q 

the  sum  is  greater  than  0,  and  we  have  only  to  show  that 
ASB  -f-  ASC  -1-  BSC  <  4  right  angles.  Produce  either 
edge,  as  AS,  to  D.  Now,  in  the  triedral  S-BCD,  BSC 
<  BSD  +  CSD.  To  each  member  of  this  inequality  add 
ASB  -I-  ASC,  and  we  have 

ASB  +  ASC  +  BSC  <  ASB  +  ASC  +  BSD  +  CSD. 
But,  ASB  4-  BSD  =  3  right  angles  (?),  and  ASC  +  CSD  - 
2  right  angles;  whence  ASB  +  ASC  -i-  BSD  ■{■  CSD  = 
4  right  angles;  and  consequently,  ASB  +  ASC  -[-  BSC  < 
4  right  angles,    q.  e.  d. 


PROPOSITION  TI. 

437  m  Theorem, — Two  iriedrals  having  the  facial  angles  of  the 
one  equal  to  the  facial  angles  of  the  other,  each  to  each.,  and  similarly 
arranged,  are  equal. 


180 


ELEMENTARY  SOLID  GEOMETRY. 


Dem.— In  the  triedrals  S  and  «,  let  ASB  =  asb,  BSC  =  bsc,  and  CSA  =  cm ; 

and  let  these  ftlcial  angles  occur  in  the  same 
order,  then  is  S-ABC  =  s-abc. 

Take  C,  any  point  in  SC,  and  make  «c=SC, 
and  draw  AC  and  BC  perpendicular  to  SC, 
and  ac  and  he  perpendicular  to  sc  ;  then  ACB 
measuues  tlie  diedral  A-SC-B,  and  ach  the  die- 
dral  a-scb.  Now  the  triangles  SCB  and  sch 
are  mutually  equiangular  (?)  and  have  SC  = 
SC  ;  hence  SB  =  ab^  and  CB  =  cb.  For  a  like 
reason  AC  =  ac,  and  SA  =  sa.  Hence  AB  — 
ab  (?),  and  the  triangles  ACB  and  acb  are  equal  (?).  Now,  since  angle  ACB, 
measuring  the  diedral  A-SC-B,  equals  angle  cicb,  measuring  the  diedral  a-8C-b, 
these  diedrals  are  equal,  and  can  be  applied  to  each  other.  Applying  these 
diedrals,  since  angle  ASC  =  asc,  and  BSC  =  bsc,  the  edges  AS  and  as  coincide, 
as  also  do  BS  and  bs,  whence  the  triedrals  coincide  throughout,  and  are  conse- 
quently equal,     q.  e.  d. 


Fio.  279. 


y^"^  PROPOSITION  yn. 

438.  Theorem, —  Of  tivo  s^ipplementary  triedrals,  the  facial 
angles  of  the  one  are  the  supplements  of  the  diedrals  of  the  other, 

Dem.— Let  S-ABC  be  any  triedral ;  if  a  second  triedral  be  foi-med  with  its 

edges  perpendicular  to  the  faces  of 
S-ABC,  one  to  each  face  respect- 
ively, then  are  the  facial  angles  of 
the  one,  supplements  of  the  die- 
drals of  the  other. 

If  the  vertex  of  the  second  trie- 
dral is  not  at  the  vertex  of  the  firet, 
we  may  conceive  a  triedral  formed 
by  drawing  three  lines  tlirough  the 
vertex  S,  as  SE,  SD,  and  SF,  paral- 
lel to  the  edges  of  the  given  trie- 
dral ;  then  will  these  edges  be  per- 
pendicular to  the  same  planes  as 
the  edges  to  which  they  are  paral- 
lel {405),   and    hence   the  angle 
thus  formed  (S-EDF)  will  be  a  triedral    supplemental  to  S-ABC  {433\  and 
the  facial  angles  of  the  two  having  their  edges  parallel  will  be  equal  {416),  and 
consequently  the  triedrals  equal  {437)-    Now,  SE  being  perpendicular  to  ASB, 
and  SF  to  ASC,  angle  ESF  is  supplemental  to  the  diedral  B-SA-C  {429.)    In 
like  manner,  S3  Lcing  perpendicular  to  BSC/DSE  is  supplemental  to  A-SB-C, 
and  DSF  to  A-SC-B. 

Thus  we  have  shown  that  the  facial  angles  of  S-EDF  are  the  supplements 


Fig.  280. 


OF    TEIEDRALS. 


181 


of  the  diedrals  of  S-ABC.  We  are  now  to  show  that  the  facial  angles  of 
S-ABC  are  supplements  of  the  diedrals  of  S-EDF ;  i.  s.,  that  ASB  is  the  sup- 
plement of  D-SE-F,  BSC  of  E-SD-F,  and  ASC  of  D-SF-E.  Since  SE  is  by  hy- 
pothesis perpendicular  to  ASB,  it  is  perpendicular  to  AS  (387)  ;  and  since  SF 
is  perpendicular  to  ASC,  it  is  perpendicular  to  AS  {387).  Hence  AS  is  per- 
pendicular to  the  face  FSE  (?).  In  like  manner  we  may  show  that  SB  is  per- 
pendicular to  DSE,  and  SC  to  DSF  ;  whence  it  follows  from  the  preceding  part 
of  the  demonstration,  or  directly  from  {429),  that  angle  ASB  is  the  supple- 
ment of  D-SE-F,  BSC  of  E-SD-F,  and  ASC  of  D-SF-E. 

438a.  SCH.— If  any  edge  of  S-EDF,  as  OS,  is  produced  beyond  S,  another 
triedral  is  formed  which  has  its  edges  perpendicular  to  the  faces  of  S-ABC. 
Thus  in  all  4  triedrals  can  be  formed  with  their  edges  perpendicular  to  the  faces 
of  S-ABC  ;  but  the  proposition  holds  only  for  S-EDF. 


PROPOSITION  nil. 

439,  TJieorem, — The  stem  of  the  diedrals  of  a  triedral  mai/  be 
anything  leiween  two  and  six  right  angles. 

Dem.— Each  diedral  being  the  supplement  of  a  plane  angle  of  the  supple- 
mentary triedral,  the  sum  of  the  three  diedrals  is  3  times  2  riglit  angles,  or 
6  right  angles  —  the  sum  of  tJie  angles  of  the  supplementary  triedral  But  this  latter 
sum  may  be  anything  between  0  and  4  right  angles  (?).  Hence  the  sum  of  the 
diedrals  may  be  anything  between  3  and  6  right  angles,    q.  e.  d. 


PROPOSITION  IX. 

440.  Theorem, — Aii  isosceles  triedral  and  its  sym- 
metrical triedral  are  eqiial. 

Dem. — Let  S-ABC  be  an  isosceles  triedral  with  the  facial  angle 
ASB  =  BSC ;  then  is  it  equal  to  its  symmetrical  triedral  S-abc. 

For,  revolve  S-abc  about  S  until  Sb  falls  in  SB,  and  bring  the 
plane  Sba  into  the  plane  SBC ;  then,  since  the  diedrals  CSB-A  and 
a-Sb-c  are  opposite,  they  are  equal  {425)*  and  the  plane  Sbc  will 
Fall  in  SBA.  Moreover,  Sa  will  fall  in  SC,  since  angle  BSC  =  ASB 
(by  hypothesis)  =  bSa  (vertical  to  ASB).  In  like  manner  Sc  will 
fall  in  SA,  and  the  triedrals  will  coincide,  and  are  therefore  equal. 

Q.  E.  D. 

441.  ScH.— If  angle  ASB  is  not  equal  to  BSC,  it  is  easy  to  see  that  the  ap 


*  Should  the  papil  have  difficulty  in  perceiving  this,  let  him  notice  that  CSB  and  cSb  art 
parts  of  one  and  the  same  plane  ;  and  ASB  and  aSb  are  parts  of  another.  Now  6B  is  tbe  inter, 
eection  of  these  planes,  and  the  diedrals  mentioned  are  on  opposite  sides  of  this  line  of  inten 
eection. 


182 


ELEMENTARY  SOLID  GEOMETRY. 


plication  will  fail,  notwithstanding  the  diedrals  are  equal,   and  the  triedraLs 
symmetrical. 

442.  Cor.  1. — The  diedrals  opposite  the  equal  facial  angles  of  an 
isosceles  triedral  are  equal. 

The  diedral  b-Sa-c  =  B-SAC being  opposite,  and  b-Sa-c  —  B-SC-A  as  shown 
in  the  demonstration  ;  hence  B-SA-C  =  B-SC-A. 

443,  Cor.  2. — Conversely,  If  two  diedrals  of  a  triedral  are  equal, 
the  triedral  is  isosceles. 

Dem.— If  B-SA-C  =  B-SC-A,  SABC  is  isosceles.  For,  revolving  S-ahc  as  be- 
fore till  the  facial  angle  aSc  falls  in  its  equal  (?)  ASC,  since  the  diedral  B-SC-A 
=  B-SA-C  (by  hypothesis)  and  the  latter  equals  its  opposite  h-Sa-c,  the  plane  hSa 
will  fall  in  the  plane  BSC  ;  and,  for  like  reasons,  the  plane  hSc  will  fall  in  BSA. 
Now,  as  these  planes  coincide,  their  intersections  Sh  and  SB  coincide,  and  the 
triedrals  are  equal ;  and  the  facial  angle  BSC  =  hSa.  But  &Sa  =  ASB  (?);  hence 
ASB  =  BSC  ;  i.  e.,  the  triedral  S-ABC  is  isosceles. 


444,  TJieoreni, 

d 


PROPOSITION  X. 

—Two  symmetrical  triedrals  are  equivalent. 

Dem.— S-ABC   is   equivalent    to  its  symmetrical 

triedral  S-dbc. 

For,  let  dD  be  so  drawn  that  the  angles  DSA, 
DSC,  and  DSB  shall  be  equal,  and  consequently  dSa 
=  dSc  =  dSb,  and  the  latter  respectively  equal  to  the 
former.  Then  the  isosceles  •  symmeti"ical  triedral 
S-DCB  =  S-dcb,  S-DCA  =  S-dea,  and  S-ADB  =  S-adb. 
Whence  the  polyedral  S-ABCD  =  S-abcd.  Now, 
from  the  former  subtracting  S-ADB,  and  from  the 
latter  S-adb,  there  remains  S-ABC  =  S-abc.    Q.  e.  d. 

445,  ScH.— If  dD  fell  within  the  given  triedrals, 
they  would  be  made  up  of  the  three  equal  isosceles 
triedrals,  and  hence  equivalent. 


Fig.  282. 


PROPOSITION  XI. 

446.  TJieorem, — Tiuo  triedrals  which  have  two  facial  angles 
and  the  included  diedral  equal,  each  to  each,  are  equal,  or  symmetri- 
cal and  equivalent. 

Dem.— If  the  equal  faces  are  on  the  same  sides  of  the  diedral  in  the  two 
tiiedrals,  the  one  figure  can  be  applied  to  the  other ;  and  if  they  are  on  different 


OF  TBIEDRALS.  183 

sides,  the  edges  of  one  triedral  may  be  produced,  forming  the  symmetrical  tri- 
edral,  to  which  the  other  given  triedral  may  be  appUed.  [Let  the  student  con- 
struct figures,  and  go  through  with  the  application.] 


PROPOSITION  xn. 

447*  TJieorem, — Two  triedrals  which  have  two  diedrals  and 
the  included  facial  angle  equal,  are  equal,  or  symmetrical  and  equiva- 
lent. 

Dem. — [Same  as  in  the  preceding.  Let  the  student  draw  figures  like  those 
for  the  preceding,  and  go  through  with  the  details  of  the  application.] 

44S,  Cor. — It  loill  he  observed  that  in  equal  or  in  symmetrical 
triedrals,  the  equal  facial  angles  are  opposite  the  equal  diedrals. 


PROPOSITION  xni. 

449.  TJieorem, — Two  triedrals  which  have  tivo  facial  angles 
of  the  one  equal  to  tzvo  facial  angles  of  the  other,  each  to  each,  and  the 
included  diedrals  unequal,  have  the  third  facial  angles  unequal,  and 
the  greater  facial  a7igle  belongs  to  the  triedral  having  the  greater  in- 
cluded diedral. 

Dem. — Let  ASC    =   asc,  and  ASB   =  ash, 

while  the  diedral  C-SA-B  >  c-sa-h\  then  CSB  S                   s 

>  csh.  Jf\                   A. 
For,  make  the  diedral  C-SA-c?  =:  c-sa-b:  and  /rA  /  \\ 

taking  AS(9  =  asby  bisect  the  diedral  o-SA-B  with            /  |i  i  \  /     \  \ 

the  plane  ISA.    Draw  o\  and  oC,  and  conceive           /   jl  \  \  /      \  \ 

the  planes  oS\   and  oSC.      Now,  the  triedral  y\-v[i't;^  o/,-------^-->^ 

S-AoC  =  s-abc,  since  they  have  two  facial  angles  /    ^:f^-\o    ^  '         i\ 

and  the  included  diedral  equal  {446).    For  a  i      i 

like  reason  S-AI(?  =  S-AIB,  and  the  facial  angle                        „  „^„ 

Fio  2S3 
oS\  =  ISB.    Again,  in  the  triedral  S-bC,  oSI  + 

ISC  >  oSC  {434),  and  substituting  ISB  for  oS\,  we  have  ISB  +  ISC  (or  BSC)  > 

oSC,  or  its  equal  bsc.    q.  e.  d. 

450.  Cor. — Conversely,  If  the  tico  facial  angles  are  equal,  each  to 
each,  in  the  ttvo  triedrals,  and  the  third  facial  angles  imequal,  the 
diedral  opposite  the  greater  facial  angle  is  the  greater. 

That  is,  if  ASB  =  ash,  and  ASC  =  asc,  while  BSC  >  bsc,  the  diedral  B-AS-C 

>  b-as-c.  For,  if  B-AS-C  =  b-as-c,  BSC  =  bsc  {44C>\  and  if  B-AS-C  <  b-as-c, 
BSC  <  bsc,  by  the  proposition.  Therefore^  as  B-AS-C  cannot  be  equal  to  nor 
less  than  b-as-c,  it  must  be  greater. 


184 


ELEMEXTABY  SOLED   GEOMETRY. 


PROPOSITION  xiy. 

451.  Tlieorem. — Two  triedrals  which  have  the  three  facial 
angles  of  the  one  equal  to  the  three  facial  angles  of  the  other ^  each  to 
each,  are  equal,  or  symmetrical  and  equivalent. 

Dem.— Let  A,  B,  and  C  represent  the  facial  angles  of  one,  and  a,  5,  and  c  the 
coirespondmg  facial  angles  of  the  other.  If  A  = 
rt,  B  =  &,  and  C  =  c,  the  triedrals  are  equal.  For  A 
being  equal  to  a,  and  B  to  h,  if,  of  their  included  die- 
drals,  SM  were  greater  thau.«?«,  C  would  be  greater 
thanc;  andifdiedral  SM  were  less  than  diedral 
sm,  C  would  be  less  than  c,  by  the  last  corollary. 
Hence,  as  diedral  SM  can  neither  be  greater  nor 
less  than  diedral  sm,  it  must  be  equal  to  it.  For  like 
reasons,  diedral  SH  =  diedral  sn,  and  diedral  SO 
=  diedral  so.  Therefore,  the  triedrals  are  equal, 
or  syinmetrical,  according  to  the  arrangement  of  the  faces.  Thus,  if  SN  and  m 
are  both  considered  as  lying  on  the  same  side  of  the  planes  MSO  and  mso,  the 
triedrals  are  equal ;  but,  if  one  lies  on  one  side  and  the  other  on  the  opposite 
side  of  those  planes  (SN  in  front,  and  sn  behind,  for  example),  the  diedrals  are 
symmetrical,  and  hence  equivalent 


Fig.  2S4. 


PROPOSITION  XT. 

4:52 1  Theorem. — Tioo  triedrals  which  have  the  three  diedrals 
of  the  one  equal  to  the  three  diedrals  of  the  other,  each  to  each,  are 
equ<th  or  symmetrical  and  equivalent. 

Dem. — In  the  two  triedrals  supplementary  to  the  given  triedrals,  the  facial 
angles  of  the  one  will  be  equal  to  the  facial  angles  of  the  other,  each  to  each, 
since  they  are  supplements  of  equal  diedrals  (438).  Hence,  the  supplementary 
triedrals  are  equal  or  equivalent,  by  the  last  proposition.  Now,  the  facial  angles 
of  the  first  triedrals  are  supplements  of  the  diedrals  of  the  supplementary  ; 
whence  the  corresponding  facial  angles,  being  the  supplements  of  equal  diedrals, 
are  equal.  Therefore,  the  proposed  triedrals  have  their  facial  angles  equal,  each 
to  each,  and  are  consequently  equal,  or  symmetrical  and  equivalent.     Q.  E.  D. 

453.  CoE. — All  trirectangular  triedrals  are  equal, 

454.  ScH. — The  proof  that  two  forms  are  equal,  includes  the  fact  that  cor- 
responding parts  are  equaL 


OF  POLYEDRALS. 


185 


OF  POLYEDRALS. 

4S3*  A  Convex  I^olyedral  is  a  polyedral  in  which  none  of 
the  faces,  when  produced,  can  enter  the  solid  angle.  A  section  of 
such  a  polyedral  made  by  a  plane  cutting  all  its  edges  is  a  convex 
polygon.     [See  Fig.  285.] 


PROPOSITION  XYI. 

456.  Theore^n^ — Tlie  sum  of  tlie  facial  angles  of  any  convex 
polyedral  is  less  than  four  right  angles. 

Dem. — Let  S  be  the  vertex  of  any  convex  polyedral.  Let  the  edges  of  this 
polyedral  be  cut  by  any  plane,  as  ABCDE,  which  section 
will  be  a  convex  polygon,  since  the  polyedral  is  convex. 
From  any  point  within  this  polygon,  as  0,  draw  lines  to 
its  vertices,  as  OA,  OB,  0  C,  etc.  There  will  thus  be  formed 
two  sets  of  triangles,  one  with  their  vertices  at  S,  and 
the  other  with  their  vertices  at  0 ;  and  there  will  be  an 
equal  number  in  each  set,  for  the  sides  of  the  polygon 
form  the  bases  of  both  sets.  Now,  the  sum  of  the  angles 
of  one  set  of  these  triangles  is  equal  to  the  sum  of  the 
angles  of  the  other  set.  But  the  sum  of  the  angles  at  the 
bases  of  the  triangles  having  their  vertices  at  S  is  greater 
than  the  sum  of  the  angles  at  the  bases  of  the  triangles 
having  their  vertices  at  0,   since    SBA  4- SBC>  ABC, 

SOB  +  SOD  >  BCD,  etc.  {4=34).     Therefore  the  sum  of  the  angles  at  S  is  less 
than  the  sum  of  the  angles  at  0,  i.  e.,  less  than  4  right  angles.     Q.  e.  d. 


Fio.  285. 


EXERCISES. 

1.  I  have  an  iron  block  whose  corners 
are  all  square  (edges  i-ight  diedrals,  and 
the  vertices  tri rectangular,  or  right,  trie- 
'drals).  If  I  bend  a  wire  square  around 
one  of  its  edges,  as  cS'd,  at  what  angle  do 
I  bend  the  wire  ?  If  I  bend  a  wire  ob- 
liquely around  the  edge,  as  asby  at  what 
angle  can  I  bend  it?  If  I  bend  it  ob- 
liquely, as  eS'f  at  what  angle  can  I  bend 
it? 


\^; 

i 


S' 


y 


FiQ.  -.ifeO. 


186  ELEMENTARY  SOLID  GEOMETRY. 

2.  Fig.  286  represents  the  appearance  of  a  rectangular  parallelepi- 
ped, as  seen  from  a  certain  position.  Now,  all  the  angles  of  such  a 
solid  are  right  angles :  why  is  it  that  they  nearly  all  appear  oblique  ? 
Can  you  see  a  right  parallelopiped  from  such  a  position  that  all  the 
angles  seen  shall  appear  as  right  angles? 

3.  The  diedral  angles  of  crystals  are  measured  with  great  care,  in 
order  to  determine  the  substance  of  which  the  crystals  consist.  How 
must  the  measure  be  taken  ?  If  we  measure  obliquely  around  the 
edge,  shall  we  get  the  true  value  of  the  angle  ? 

4.  Cut  out  any  triedral  from  a  block  of  wood  (or  a  potato),  and 
stick  three  pins  into  it,  as  near  the  vertex  as  you  can,  one  iu  each 
face,  and  perpendicular  to  that  face.  What  figure  do  the  three  pins 
form  ?  What  relation  does  the  angle  included  between  any  two  ad- 
jacent pins  sustain  to  one  of  the  diedrals  of  the  block  ?  Which  ones 
are  they  that  sustain  this  relation  ? 

5.  Can  three  planes  intersect  each  other  and  yet  not  form  a  trie- 
dral angle  ?  In  hoAv  many  ways  ?  Can  they  all  three  have  a  common 
point,  and  yet  not  form  a  triedral  ? 

6.  From  a  piece  of  pasteboard  cut  two  figures 
of  the  same  size,  like  ABCDS  and  alcds.  Then 
drawing  SB  and  SC  so  as  to  make  1  the  largest 
angle  and  3  the  smallest,  cut  the  pasteboard 
almost  through  in  these  lines,  so  that  it  will 
readily  bend  in  them.  Now  fold  the  edges  AS 
and  DS  together,  and  a  triedral  will  be  formed. 
From  the  piece  leads  form  a  triedral  in  like 
manner,  only  let  the  lines  sc  and  8a  be  drawn 
so  as  to  make  the  angles  1,  2,  and  3  of  the  same 
size  as  before,  while  they  occur  in  the  order 
Pig.  267.  given  in  heads.    Now,  see  if  you  can  slip  one 

triedral  into  the  other,  so  that  they  will  fit.     What  is  the  diffi- 
culty ? 

7.  In  the  last  case,  if  1  equals  I  of  a  right  angle,  2  =  ^  of  a  right 
angle,  and  3  =  f  of  a  right  angle,  can  you  form  the  triedral  ?  Why  ? 
If  you  keep  increasing  the  size  of  1,  2,  and  3,  until  the  sum  becomes 
equal  to  4  right  angles,  will  it  always  be  possible  to  form  a  triedral  ? 
How  is  it  when  the  sum  equals  4  right  angles  ? 


OF  PRISMS  AND   CYLINDERS. 


187 


SECTION  III, 


OF  PRISMS  AND  CYLINDERS. 


4:57*  A  JPrisni  is  a  solid,  two  of  whose  faces  are  equal,  parallel 
polygons,  while  the  other  faces  are  parallelograms.  The  equal  par- 
allel polygons  are  the  Bases,  and  the  parallelograms  make  up  the 
Lateral  or  Convex  Surface.  Prisms  are  triangular,  quadrangular, 
pentagonal,  etc.,  according  to  the  number  of  sides  of  the  polygon 
forming  a  base. 

4a8.  A  Hight  I*rism  is  a  prism  whose  lateral  edges  are  per- 
pendicular to  its  bases.  An  Oblique  prism  is  a  prism  whose  late- 
ral edges  are  oblique  to  its  bases. 

459,  A  ^Regular  Prism  is  a  right  prism  whose  bases  are 
regular  polygons ;  whence  its  faces  are  equal  rectangles. 

460,  The  Altitude  of  a  prism  is  the  perpendicular  distance 
between  its  bases  :  the  altitude  of  a  right  prism  is  equal  to  any  one 
of  its  lateral  edges. 

461,  A  Truncated  JPrism  is  a  portion  of  a  prism  cut  off  by 
a  plane  not  parallel  to  its  base.  A  section  of  a  prism  made  by  a  plane 
perpendicular   to   its   lateral 

edges  is  called  a  Right  Section 

Ill's. — In  the  figure,  (a) 
and  (6)  are  both  prisms : 
{a)  is  oblique  and  (6)  right, 
PO  represents  the  altitude 
of  (a);  and  any  edge  of 
(ft),  as  5B,  is  its  altitude. 
ABCDEF,  and  ahcdef,  are 
lower  and  upper  bases, 
respectively.  Either  por- 
tion of  ip)  cut  ojff  by  an 
oblique  plane,  as  a'h'c'd'e\ 
is  a  truncated  prism. 

462,  A  I^arallelopiped  is  a  prism  whose  bases  are  parallel- 
ograms :  its  faces,  inclusive  of  the  bases,  are  consequently  all  parallel- 


FiG.  28 


188 


ELEMENTARY  SOLID  GEOMETRY. 


ograms.    If  its  faces  are  all  rectangular,  it  is  a  rectangular  parallel- 
epiped. 

463,  A  Cube  is  a  rectangular  parallelepiped  whose  faces  are  all 
equal  squares. 


PROPOSITION  I. 

40 4*  Theorem, — Parallel  plane  sections 
of  any  prism  are  equal  polygons. 

Dem.— Let  ABCDE  and  cibcde  be  parallel  sections  of 
the  prism  MN  ;  then  are  they  equal  polygons. 

For,  the  intersections  with  the  lateral  faces,  as  ah 
and  AB,  etc.,  are  parallel,  since  they  are  intersections 
of  parallel  planes  by  a  third  plane  {410).  Moreover, 
these  intersections  are  equal,  that  is,  ab  =  AB,  be  =  BC, 
cd  =  CD,  etc.,  since  they  are  parallels  included  between 
parallels  (242).  Again,  the  corresponding  angles  of 
these  polygons  are  equal,  that  is,  a  =  A,  6  =  B,  c  =  C, 
etc.,  since  their  sides  are  parallel  and  lie  in  the  same 
direction  {416).  Therefore  the  polygons  ABCDE,  and 
abcde,  are  mutually  equilateral  and  equiangular;  that 
is,  they  are  equal.     <^.  e.  d. 

46S,  Cor. — Any  plane  section  of  apristn,  parallel  to  its  base,  is 
equal  to  the  base  ;  and  all  right  sections  are  equal. 


Fig.  289. 


PROPOSITION   IL 

466»  Theorem, — If  three  faces  including  any  triedral  of  one 
prism  are  equal  respectively  to  three  faces  including  a  triedral  of  the 
other,  and  similarly  placed,  the  prisms  are  equal, 

Dem. — In  the  prisms  AfZ,  and  k'd', 
let  ABCDE  equal  A'B'C'D'E',  fKBba  = 
A'B'b'a',  and  BCcb  =  B'C'c'b';  then  are 
the  prisms  equal. 

For,  since  the  facial  angles  of  the 
triedrals  B  and  B'  are  equal,  the  trie- 
drals  are  equal  {451),  and  being  ap- 
plied they  will  coincide.  Now,  con- 
ceiving A'rf'  as  applied  to  Arf,  with  B' 
in  B,  since  the  bases  are  equal  poly- 
gons, they  will  coincide  throughout ; 
and  the  faces  aB  and  a'B'  will  also 
coincide.     Whence,  as  a'b'  falls  in  a&, 


OF  PRISMS. 


189 


and  Vc'  in  he,  the  upper  bases,  which  are  equal  because  equal  to  the  equal  lower 
bases,  will  coincide.  Therefore  the  remaining  edges  will  have  two  points  com- 
mon in  each,  and  will  consequently  coincide. 

467*  Cor.  1. —  Two  right  prisms  having  equal  bases  and  equal 
altitudes  are  equal. 

If  the  faces  are  not  similarly  arranged,  one  prism  can  be  inverted. 

4:68,    Cor.  2. — The  above  proposition  applies  also  to  truncated 
prisms. 


PROPOSITION  m. 

469,  Tlieorem* — Any  oblique  prism  is 
equivalent  to  a  right  prism,  whose  bases  are  right 
sections  of  the  oblique  prism,  and  whose  edge  is 
equal  to  the  edge  of  the  oblique  prism. 

Dem. — Let  LB  be  an  oblique  prism,  of  which  abcde  and 
fghil  are  right  sections,  and  gh  —  OB-^  then  is  lb  equiva- 
lent to  LB.  For  the  truncated  prisms  10  and  eB  have  the 
faces  including  any  triedral,  as  C  and  B,  equal  and  simi- 
larly placed  (?),  whence  these  prisms  are  equal  {466). 
Now,  from  the  whole  figure,  take  away  prism  10,  and 
there  remains  the  oblique  prism  LB  ;  also,  from  the  whole 
take  away  the  prism  eB,  and  there  remains  the  right 
prism  Jb.  Therefore,  the  right  prism  lb  is  equivalent  to 
the  oblique  prism  LB.    q.  e.  d. 


PROPOSITION  IT. 

■The  opposite  faces  of  a  parallelopiped  are 


470,  Theorem. 

equal  and  parallel. 

Dem. — Let.  Ac  be  a  parallelopiped,  AC  and  ac  being 
its  equal  bases  (462) ;  then  are  its  opposite  faces  equal 
'  and  parallel. 

Since  the  bases  are  parallelograms,  AB  is  equal  and 
parallel  to  DC  ;  and,  since  the  faces  are  parallelograms, 
aA  is  equal  and  parallel  to  dD.  Hence  angle  «AB  = 
e?DC,  and  their  planes  are  parallel,  since  their  sides  are 
parallel  and  extend  in  the  same  directions.  Therefore 
aB  and  dO  are  equal  {301)  and  parallel  parallelograms. 
In  like  manner  it  may  be  shown  that  aD  is  equal  and 
parallel  to  bO. 


190 


ELEMENTARY  SOLID  GEOMETRY. 


PROPOSITION  Y. 

4:71*    Theorem^ — Tlie  diagonals  of  a  parallelopiped  bisect  each 
other, 

Dem. — Pass  a  plane  through  two  opposite  edges, 
as  &B  and  dD.  Since  the  bases  are  parallel,  hd  and 
BD  will  be  parallel  {410),  and  bBDd  will  be  a  paral- 
lelogram. Hence,  bD  and  dB  are  bisected  at  o  (?). 
For  a  like  reason,  passing  a  plane  through  dc  and  AB, 
we  may  show  that  dB  and  cA  bisect  each  other,  and 
hence  that  cA  passes  through  the  common  centre  of 
dB  and  bD.  So  also  aC  is  bisected  by  6D,  as  appears 
from  passing  a  plane  through  ab  and  DC.  Hence,  all 
the  diagonals  are  bisected  at  o.    q,.  e.  d. 

Fig.  293. 

4  72,  Cor. — Tlie  diagonals  of  a  rectangular  parallelopiped  are  equal. 


PROPOSITION  TL 

47 3 •  Theorem, — A  parallelopiped  is  divided  into  ttvo  equiva- 
lent triangular  prisms  hy  a  plane  passing  through  its  diagonally 
opposite  edges. 

Dem.— Let  H-ABCD  be  a  parallelopiped,  divided 
through  its  diagonally  opposite  edges  FA  and  HC; 
then  are  the  triangulai*  prisms  H-ABC,  and  L-ADC 
equivalent. 

For  this  parallelopiped  is  equivalent  to  a  right 
parallelopiped  having  a  right  section  N}cd  for  its  base, 
and  AF  for  its  edge  {469),  i.  e.,  H-ABCD  is  equiva- 
lent to  h-fiibcd.  For  the  same  reason  the  oblique 
triangular  prism  H-ABC  is  equivalent  to  the  right 
triangular  prism  h-Abc;  and  L-ADC  is  equivalent 
to  l-Mc.  But  Ji-Mc  is  equal  to  h-Mc,  as  they 
are  right  prisms  with  equal  bases  {467)  and  a  com- 
mon altitude.  Hence,  H-ABC  is  equivalent  to  L-ADC, 
they  are  equivalent  to  two  equal  prisms,    q.  e.  d. 


'^ 


Fig.  294. 


PROPOSITION  yn. 

474,  Theoretn, — Any  parallelopiped  is  equivalent  to  a  redan' 
gular  parallelopiped  having  an  eq\dvalent  base  and  the  same  altitude. 


OF  PRISMS. 


191 


Dem. — Let  H-ABCD  be  any  parallelopiped 
with  all  its  faces  oblique.  1st.  By  making  the 
right  section  rtclHe,  and  completing  the  paral- 
lelepiped adHebcCf,  we  have  an  equivalent  right 
parallelopiped  (409).  2d.  Through  the  edge 
ef  of  this  right  parallelopiped  make  the  right 
section  ea'b'f  and  complete  the  parallelopiped 
ea'b'/Hd'c'C,  and  we  have  a  rectangular  paral- 
lelopiped equivalent  to  the  one  previously 
formed  {469),  and  hence  equivalent  to  the 
given  one.  Now,  the  base  of  this  rectangular 
parallelopiped,  i.  e.,  a'b'c'd',  is  equal  to  abed  (?), 
which  in  turn  is  equivalent  to  ABCD  (?).  Moreover,  the  altitude  of  the 
rectangular  parallelopiped  is  the  same  as  that  of  the  given  one,  since  their 
bases  lie  in  the  same  parallel  planes  Ac'  and  EG.  Therefore,  the  parallelopiped 
H-ABCD  is  equivalent  to  the  rectangular  parallelopiped  H-a'b'c'd',  which 
has  an  equivalent  base  and  the  same  altitude,    q.  e.  d. 


A 

A 

Jl 

/ 

t\ 

Ll\  V 

/ 

k 

//i'u^_ 

// 

1/  ■»' 

B 

295. 

"i/b 

a 

Fig. 

•6 

PROPOSITION  vm.  \D 

47S,  Theorem, — The  area  of  the  lateral  surface  of  a  right 
prism  is  equal  to  the  product  of  its  altitude  into  the  perimeter  of 
its  base. 

I)em. — The  lateral  faces  are  all  rectangles,  having  for  their  common  alti- 
tude the  altitude  of  the  prism  {460).  Whence  the  area  of  any  face  is  the 
product  of  the  altitude  into  the  side  of  the  base  which 
forms  its  base ;  and  the  sum  of  the  areas  of  the  faces 
is  the  common  altitude  into  the  sum  of  the  bases  of  /  /  /  /  / 
the  faces,  that  is,  into  the  perimeter  of  the  base  of  /  /  /  •'  •'  -^ 
Ihe  prism,    q.  e.  d.  /  /  .g . 


.  476.  A  Cylindrical  Surface  is  a 

curved  surface  traced  by  a  straight  line  moving 
so  as  to  remain  constantly  parallel  to  its  first 
position,  while  any  point  in  it  traces  some 
curve.  The  moving  line  is  called  the  Gener- 
atrix, and  the  curve  traced  by  a  point  of  the 
line  is  the  Directrix. 


Fig.  296. 


III. — Suppose  a  line  to  start  from  the  position  AB.  and  move  towards  N  in 


192 


ELEMENTARY  SOLID  GEOMETRY. 


such  a  manner  as  to  remain   all  the  time  parallel  to   its   first  position  AB, 

whUe  A  traces  the  curve  A123456 M.    The  surface  thus  traced  is  a  Gyliil 

drical  Surface;  AB  is  the  Geiierattix,  and  the  curve  ANM  the  Directrix. 

477.  A  Circular  Cylinder,  called  also  a  Cylinder  of  Revolu- 
tion, is  a  solid  generated  by  the  revolution  of  a  rectangle  around  one 
of  its  sides  as  an  axis. 


B" 


III.— Let  COAB  be  a  rectangle,  and  conceive  it  re- 
volved about  CO  as  an  axis,  talviDg  successively  the 
positions  COA'B',  COA"B",  etc. ;  the  solid  generated  is  a 
Circular  Cylinder,  or  a  cylinder  of  revolution.  Tlie  re- 
volving side  AB  is  the  generatrix  of  the  surface,  and 
the  circumference  OA  (or  CB)  is  the  directrix.  This  is 
tlie  only  cylinder  treated  in  Elementary  Geometry,  and 
is  usually  meant  when  the  word  Cylinder  is  used  without 
specifying  the  kind  of  cylinder. 

478,  Tlie  Axis  of  the  cylinder  is  the  fixed 
side  of  the  rectangle.  The  side  of  the  rectangle 
opposite  the  axis  generates  the  Convex  Surface ; 
while  the  other  sides  of  the  rectangle,  as  OA  and 
CB,  generate  the  Bases,  which  in  the  cylinder  of 
revolution  are  circles.  Any  line  of  the  surface  corresponding  to 
some  position  of  the  generatrix  is  called  an  Element  of  the  surface. 

479.  A  Right  Cylinder  is  one  whose  elements  are  perpen- 
dicular to  its  base.  In  such  a  cylinder  any  element  is  equal  to  the 
ftxis.     A  Cylinder  of  Revolution  {477)  is  right. 

480,  A  prism  is  said  to  be  inscribed  in  a  cylinder,  when  the  bases 
of  the  prism  are  inscribed  in  the  bases  of  the  cylinder,  and  the  edges 
of  the  prism  coincide  with  elements  of  the  cylinder. 


PROPOSITION  IX. 

481,  Theorem. —  The  area  of  the  convex  surface  of  a  cylinde? 
of  revolution  is  equal  to  the  product  of  its  axis  into  the  circumfer' 
ence  of  its  base,  i.  e.,  2;rRH,  H  bei7ig  the  axis  and  E  the  radius  of 
the  base. 


OP  PRISMS  AND   CYLINDERS. 


193 


Dem.— Let  a  right  prism,  with  any  regular  polygon  for 
its  base,  be  inscribed  in  the  cylinder,  as  k-abcdef,  in  the 
cylinder  whose  axis  is  HO.  The  area  of  the  lateral  surface 
of  the  prism  is  HO  (=  hb)  into  the  perimeter  of  its  base, 
i.e.,  HO  k  {ah  +  hc  +  cd  +  de  +  ef  +  fa).  Now,  bisect  the 
arcs  ah,  he,  etc.,  and  inscribe  a  regular  polygon  of  twice  the 
number  of  sides  of  the  preceding,  and  on  this  polygon  as 
a  base  construct  the  right  inscribed  prism  with  double  the 
number  of  faces  that  the  first  had.  The  area  of  the  lateral 
surface  of  this  prism  is  HO  x  the  perimeter  of  its  base.  In 
like  manner  conceive  the  operation  of  inscribing  right 
prisms  with  regular  polygonal  bases  continually  repeated ; 
it  will  ahoays  be  true  that  the  area  of  the  lateral  surface 
is  equal  to  HO  x  t?ie  perimeter  of  the  base.  But  the  circum- 
ference of  the  base  of  the  cylinder  is  the  limit  toward 
which  the  perimeters  of  the  inscribed  polygons  forming  the  bases  of  the  prisms 
constantly  approach,  and  the  convex  surface  of  the  cylinder  is  the  limit  of  the 
lateral  surface  of  the  inscribed  prism.  Therefore,  the  area  of  the  convex  sur- 
face of  the  cylinder  is  HO  into  the  circumference  of  the  base.  Finally,  if  R  i» 
the  radius  of  the  base,  2;rR  is  its  circumference.  This  multiplied  by  H  th» 
altitude,  i.  e.,  H  x  27rR,  or  27rRH,  is  the  area  of  the  convex  surface  of  the 
cylinder. 


Fig.  298. 


PROPOSITION  X. 

4:82,  Tlieoj^ein. — The  volume  of  a  rectangtdar  parallelopiped 
is  equal  to  the  product  of  the  three  edges  of  one  of  its  triedrals. 


1^ 


Dem. — Let  H-CBFE  be  a  rectangular  paral- 
lelopiped. 1st.  Suppose  the  edges  commen- 
surable, and  let  BC  be  5  units  in  length,  BA 
4,  and  BF  7.  Now  conceive  a  cube,  as  d-fbBg 
whose  edge  is  one  of  these  linear  units.  This 
cube  may  be  used  as  the  unit  of  volume.  Con- 
ceive the  parallelopiped  0-caBb,  whose  length 
is  7,  and  whose  edges  ca  and  cb  are  1  (the 
linear  unit  of  measure  assumed).  This  paral- 
lelopiped will  contain  as  many  of  the  units 
of  volume  as  there  are  linear  units  in  BF: 
we  suppose  7.  Again,  conceive  the  paral- 
lelopiped whose  base  is  EGBF  and  altitude  PE,  one  of  the  linear  units.  This 
parallelopiped  will  contain  as  many  of  the  former  as  there  nrc  linear  units  in 
BC:  we  suppose  5.    Hence  this  last  volume  is  5  x  7  =  35.    Finally,  there  will 

13 


FiH.  29!). 


194  ELEMENTARY   SOLID  GEOMETRY. 

be  as  many  times  this  number  of  imits  of  volume  in  the  whole  parallelepiped 
as  AB  contains  linear  units,  or  4  x  35  =  140.  Hence,  when  the  edges  are 
commensurable,  the  volume  is  the  product  of  the  three  edges  including  a 
triedral. 

2nd.  Wlien  the  edges  are  not  commensurable,  we  reach  the  same  conclusion 
by  takini;  successively  a  smaller  and  smaller  linear  unit.  Thus,  for  a  first 
approximation  take  some  aliquot  part  of  one  edge,  as  i^g  of  FB.  Now,  by  hypo- 
thesis this  is  not  contained  an  exact  number  of  times  in  BC,  nor  in  BA.  But 
conceive  it  as  applied  to  BC  as  many  times  as  it  can  be  ;  the  remainder  will  be 
less  than  -,^  FB.  In  like  manner  conceive  it  applied  to  AB.  The  volume  of  the 
parallelopiped  included  by  these  edges  will  be  measured  by  the  product  of  the 
edges.  Now  conceive  the  linear  unit  smaller.  The  unmeasured  portion  will 
be  less.  Thus,  b}^  supposing  the  linear  unit  to  diminish  indefinitely,  we  see  that 
it  will  cdways  remain  true  tliat  the  measure  is  the  product  of  the  three  edges 
forming  a  triedral. 

483,  Cor.  1. — The  volume  of  a  cuhe  is  the  third  power  of  its 
edge. 

4S4,  ScH.— This  fact  gives  rise  to  the  term  cuhe,  as  used  in  arithmetic  and 
algebra,  for  "  third  power." 

485.  Cor.  2. — The  volume  of  a  rectangular  parallelopiped  is 
equal  to  tlie  product  of  its  altitude  into  the  area  of  its  base,  the  linear 
unit  being  the  same  for  the  measure  of  all  the  edges. 

486.  Cor.  3. — T7ie  volume  of  a7ii/ parallelopiped  is  equal  to  the 
product  of  its  altitude  and  the  area  of  its  base. 

For  any  parallelopiped  is  equivalent  to  a  rectangular  parallelopiped  having 
an  equivalent  base  and  the  same  altitude  {474). 

487.  Cor.  4. — Parallelepipeds  of  the  same  or  equivalent  bases  are 
to  each  other  as  their  altitudes,  and  those  of  the  same  altitudes  are  to 
each  other  as  their  bases.  And,  in  general,  jmrallelopipeds  are  to 
each  other  as  the  products  of  their  bases  and  altitudes. 


PROPOSITION  XI. 

488.  Tlieorem. — The  volume  of  any  prism  is  equal  to  the  pro 
■duct  of  its  altitude  into  its  base. 


OF   PRISMS   AND   CYLINDERS. 


195 


Dem. — 1st.  Let  E-ABD  be  a  triaDgular  prism.  Com- 
plete the  parallelepiped  E-ABCD.  Then  is  E-ABD  = 
i  E-ABCD  {473).  But  the  vokime  of  E-ABCD  is 
equal  to  its  altitude  into  its  base ;  hence  the  volume 
of  E-ABO  is  equal  to  its  altitude  into  ^ABCD,  or 
ABD. 

2d.  Any  prism  may  be  divided  into  partial,  tri- 
angular prisms,  by  passing  planes  through  one  edge 
and  all  the  other  non-adjacent  edges,  as  in  the  figure. 
Let  H  be  the  altitude  of  the  whole  prism,  then  is  it 
also  the  common  altitude  of  the  partial  prisms.  Now, 
the  volume  of  each  triangular  prism  is  H  into  its 
base ;  hence,  the  sum  of  the  volumes  is  H  into  the 
sum  of  the  bases,  i.e.,  H  into  the  base  of  the  whole 
prism. 

489.  Cor.  1. — T/ie  volume  of  a  right  prism 
is  equal  to  the  product  of  its  edge  into  its 
base. 

490.  Cor.  2. — Prisms  of  the  same  altitude 
are  to  each  other  as  their  bases ;  and  prisms 
of  the  same  or  equivalent  bases  are  to  each 
other  as  their  altitudes;  and,  i?i  general, 
prisms  are  to  each  other  as  the  products  of 
their  bases  and  altitudes. 


PROPOSITION  xn. 

491,  Theorem.— Tlie  volume  of  a  cylinder  of  revolution  is 
equal  to  the  product  of  its  base  and  altitude,  i.  e.,  ;rR'H,  H  being  the 
altitude  and  K  the  radius  of  the  base. 

Dem. — Inscribe  any  regular  right  j)i-ism  in  the  cylinder, 
as  in  {4:81).  The  volume  of  this  prism  is  equal  to  the 
product  of  its  base  and  altitude ;  and  this  continues  to  be 
the  fact  as  the  number  of  sides  of  the  polygon  forming  the 
bftse  is  successively  doubled,  and  the  prism  approaches 
equality  with  the  cylinder.  Hence,  as  the  volume  of  the 
prism  is  always  equal  to  the  product  of  its  base  and  alti- 
tude, and  as  the  altitude  of  the  prism  remains  equal  to  the 
altitude  of  the  cylinder,  this  fact  is  true  when  the  number 
of  the  sides  of  the  base  of  the  prism  is  infinitely  multiplied  ; 
whence  the  volume  of  the  cylinder  is  equal  to  the  product 
of  its  base  and  altitude.  Now,  K  being  the  radius  of  the 
base,  the  area  of  the  base  is  ^rR^  (?) :  hence,  the  volume  of 
the  cylinder  is  equal  to  ;rIl^H. 


196  ELEMENTAKY  SOLID  GEOMETRY. 

492,  Cor. — Tlie  volume  of  any  cylinder  is  equal  to  the  product 
of  its  base  into  its  altitude. 

This  can  be  demonstrated  in  a  manner  altogether  analogous  to  the  case 
given  in  the  proposition. 


493.  Similar  Solids  are  such  as  have  their  corresponding 
solid  angles  equal  and  their  homologous  edges  proportional. 

494.  Similar  Cylinders  of  reyolution  are  such  as  have  their 
altitudes  in  the  same  ratio  as  the  radii  of  their  bases. 

495.  Homologous  Edges  of  similar  solids  are  such  as  are 
included  between  equal  plane  angles  in  corresponding  faces. 

Ill's.— The  idea  of  similarity  in  the  case  of  solids  is  the  same  as  in  the 
case  of  plane  figures,  viz.,  that  of  liksness  of  form.  Thus,  one  would  not  think 
such  a  cylinder  as  one  joint  of  stovepipe,  similar  to  another  composed  of  d 
hundred  joints  of  the  same  pipe.  One  would  be  long  and  very  slim  in  propor. 
tion  to  its  length,  while  the  other  would  not  be  thought  of  as  slim.  But,  if  w« 
have  two  cylinders  the  radii  of  whose  bases  are  2  and  4,  and  whose  lengths  ar« 
respectively  6  and  12,  we  readily  recognize  them  as  of  the  same  shape:  they 
are  similar. 


PROPOSITION  xin. 

496.  TJieorem^. — The  lateral  surfaces  of  similar  right  prisma 
are  to  each  other  as  the  squares  of  their  edges  (or  altitudes)  and  as 
the  squares  of  any  tiuo  homologous  sides  of  their  bases,  i.  e.,  as  tht 
squares  of  any  two  homologous  lines. 

Dem.— Let  A,B,  C,  D,  and  E,  be  the  sides  of  the  base  of  one  right  prism 
•whose  edge  (equal  to  its  altitude)  is  H,  and  a,  b,  c,  d,  and  e,  the  homologous 
sides  of  a  similar  piism  whose  edge  is  h.  Letting  A  +  B  +C  +  D+  E  =  P^ 
and  a  +  b  +  c  +  d+e  =  p,we  have 

P:p  ::  A:  a  ::B:b  ::C:c,  etc.  (?). 
But  by h3rp«tiiesis,       H  :h::  A  :  a  :  :  B  :  b,  etc. 
Hence,  A  P  :  p  :  :  H  .  h  (?). 

Now,  H:h::R:h  {?). 

Whence,  P  y.  H :  p  x  h  :  :  ET"  :h}  (?). 

And  as  5""  :  7i'  :  :  A'  lo'  :  -.B'  :b\  etc., 

we  have  P  x  H:p  x  h  ::  A^  .  a^  : :  ^  :  6',  etc. 

But  P  X  H  is  the  area  of  the  lateral  surface  of  one  prism  and  p  x  A  of  thft 
other,  whence  the  truth  of  the  theorem  appeara. 


OF  PRISMS  AND  CYLINDERS 


197 


PROPOSITION  xiy. 

49 7 •  Theorem. — The  volumes  of  similar  ^^risms  are  to  each 
other  as  the  cubes  of  their  homologous  edges,  and  as  the  cubes  of 
their  altitudes. 

Dem.— Let  H-ABCDE  and  h-ahcde  be 
two  similar  prisms,  of  which  A  and  a  are 
corresponding  triedrals.  Placing  «  so  that 
it  will  coincide  with  A,  all  the  faces  and 
edges  of  one  will  be  parallel  to  or  coinci- 
dent with  the  corresponding  parts  of  the 
other,  by  definition  {493).  Let  fall  the 
perpendicular  FP  upon  the  common  base, 
or  its  plane  produced,  so  that  FP  shall 
equal  the  altitude  of  H-ABCDE,  and  OP, 
intercepted  between  the  planes  of  the  upper 
and  lower  bases  of  h-abcde^  shall  be  its  alti- 
tude. Call  the  former  altitude  H,  and  the 
latter  h.  Since  FP  and  AF  are  cut  by- 
parallel  planes,  we  have 

AF  :  a/  : :  H  :  A;  and  AB  :  aJ  : :  H  :  h, 
since  by  definition  AF  :  af  : :  AB  :  ab^  etc. 
Call    the  base  of   H-ABCDE    B,  and  of 

Now,  as  the  bases  are  similar  polygons, 

B  :  6  : :  A"b'  :  :^'  :  :  H^  :  h\ 

H  :  /i  :  :  AB  :  a6  :  :  H  :  ^. 

B  X  H  :  6  X  7i  ::  AB' :  ^'  ::  H"  :  7i'. 
Now,  as  B  X  H  and  b  x  h  are  the  volumes  of  the  respective  prisms,  and  as 
AB    :  ah^  as  the  cubes  of  any  other  homologous  edges  are  to  each  other,  the 
truth  of  the  theorem  is  demonstrated. 


Fig.  .302. 


h-abcde  b. 

But 
Hence, 


PROPOSITION  XV. 

408,  Theorem, — The  convex  surfaces  of  similar  cylinders  of 
revolution  are  to  each  other  as  the  squares  of  their  altitudes,  and  as 
{lie  squares  of  the  radii  of  their  bases. 

Dem.— Let  H  and  h  be  the  altitudes,  and  R  and  r  the  radii  of  the  base^  of 
two  similar  C3'linders ;  the  convex  surfaces  are  2;rRH  and  27crh  (481).    Now, 
27rRH  :  27trh  : :  RH  :  r^  (?)  (1). 

By  hypothesis,  H  :  7i  : :  R  :  r,  or 


H       h 
R 


..  .R  r 
-  and  ^  =  T 
r         HA. 


Multiplying  the  terms  of  the  second  couplet  of  (1)  by  these  cqunls,  we  have, 

27rRH  :  "litrh  :  :  H^  :  ^^ 
and  2;rRH  :  2;rr74  :  :  R"'  :  r'.    Q.  E.  D. 


198  ELEMENTARY  SOLID   GEOMETRY. 

PROPOSITION  XVI. 

499.  Theorem. — The  volumes  of  similar  cylinders  of  revolu- 
tio?i  are  to  each  other  as  the  cubes  of  their  altitudes,  or  as  the  cubes 
of  the  radii  of  their  bases. 

Dem. — Using  the  same  notation  as  in  the  last  demonstration,  the  student 
should  be  able  to  give  the  reasons  for  the  following  steps. 

R  :  r  : :  H  :  A  (?),  whence  7tW  :  Tti-^  : :  W  :  A'  (?).  Multiplying  the  last 
proportion  hy  H  :  h  : :  H  :  ^,  we  have  ttR^H  :  itt^h  :  :  H*  :  h\  or  as  R'  :  ?  ^ 
since  H'  ://=*::  R"  :  7'^  (?).  Now,  ;rR''H  and  7t)'='h  are  the  volumes  of  the 
cylinders  (?) ;  hence  the  volumes  ai-e  to  each  other  as  the  cubes  of  xAe  altitudes, 
or  as  the  cubes  of  the  radii  of  the  bases,     q.  e.  d. 

ScH.— It  is  a  general  truth,  that  the  surfaces  of  similar  solids,  of  any  form,  are 
to  each  other  as  the  squares  of  homologous  lines ;  and  their  volumes  are  as  the 
cubes  of  such  lines. 


EXERCISES. 

1.  A  farmer  has  two  grain  bins  which  are  parallelepipeds.  The 
front  of  one  bin  is  a  rectangle  6  feet  long  by  4  high,  and  the  front 
of  the  other  a  rectangle  8  feet  long  by  4  high.  They  are  built 
between  parallel  walls  5  feet  apart.  The  bottom  and  ends  of  the 
first,  he  says,  are  "  square "  (he  means,  it  is  a  rectangular  parallelo- 
piped),  while  the  bottom  and  ends  of  the  other  slope,  i.  e.,  are  oblique 
to  the  front.    What  are  the  relative  capacities  of  the  bins  ? 

2.  How  many  square  feet  of  boards  in  the  walls  and  bottom  of  the 
first  bin  mentioned  in  Ex.  1? 

3.  An  average  sized  honey  bee's  cell  is  a  right  hexagonal  prism, 
.8  of  an  inch  long,  with  faces  -^  of  an  inch  wide.  The  width  of  the 
face  is  always  the  same,  but  the  length  of  the  cell  varies  according 
to  the  space  the  bee  has  to  fill.  Are  honey  bee's  cells  similar  ?  Is  a 
honey  bee's  cell  of  the  dimensions  given  above,  similar  to  a  wasp's 
cell  which  is  1.6  inches  long,  and  whose  face  is  .3  of  an  inch  wide? 
How  much  more  honey  will  the  wasp's  cell  hold  than  the  honey 
bee's?  "i/  0.- 

4.  How  many  square  inches  of  sheet-iron  d6es  it  take  to  make  a 
joint  of  7-inch  stovepipe  2  feet  4  inches  long,  allowing  an  inch  and 
a  half  for  making  the  seam  ? 

5.  A  certain  water-pipe  is  3  inches  in  diameter.  How  much  water 
is  discharged  through  it  in  24  hours,  if  the  current  flows  3  feet  per 


OF  PYRAMIDS  AND  CONES.  199 

minute  ?    How  much  through  a  pipe  of  twice  as  great  diameter,  at 
the  same  rate  of  flow  ? 

6.  What  is  the  ratio  of  the  length  of  a  hogshead  holding  125  gal- 
Ions,  to  the  length  of  a  keg  of  the  same  shape,  holding  8  gallons  ? 

7.  What  are  the  relative  amounts  of  cloth  required  to  clothe  3 
men  of  the  same  form  (similar  solids),  one  being  5  feet  high,  another 
5  feet  9  inches,  and  the  other  6  feet,  provided  they  dress  in  the  same 
style?  If  the  second  of  these  men  weighs  156  lbs.,  what  do  the 
others  weigh  ? 

8.  If  a  man  5^  feet  high  weighs  160  lbs.,  and  a  man  3  inches  taller 
weighs  180  lbs.,  which  is  the  stouter  in  proportion  to  his  height  ? 

9.  I  have  a  prismatic  piece  of  timber  from  which  I  cut  two  blocks 
both  5  feet  long  measured  along  one  edge  of  the  stick;  but  one 
block  is  made  by  cutting  the  stick  square  across  (a  right  section), 
and  the  other  by  cutting  both  ends  of  it  obliquely,  making  an  angle 
of  45°  with  the  same  face  of  the  timber.  Which  block  is  the  greater  ? 
Which  has  the  greater  lateral  surface  ? 

10.  HoAV  many  cubic  feet  in  a  log  12  feet  long  and  2  feet  5  inches 
in  diameter?  How  many  square  feet  of  inch  boards  can  be  cut 
from  such  a  log,  allowing  ^  for  waste  in  slabs  and  sawing  ? 


/ 
SECTION  IV,  \i\   ,  ;-    ,\  , 

OF  PYRAMIDS  AND  CONES.  ; 


500*  A  pyramid  is  a  solid  having  a  polygon  for  its  base, 
and  triangles  for  its  lateral  faces.  If  the  base  is  also  a  triangle,  it  is 
called  a  triangular  pyramid,  or  a  tetraedron  {i.e.,  a  solid  with  four 
faces).  The  vertex  of  the  polyedral  angle  formed  by  the  faces  is  the 
vertex  of  the  pyramid. 

501,  Tlie  Altitude  of  a  pyramid  is  the  perpendicular  dis- 
tance from  its  vertex  to  the  plane  of  its  base. 

502,  A  Might  Pyramid  is  one  whose  base  is  a  regular 


200 


ELEMENTARY  SOLID   GEOMETRY. 


polygon,  and  the  perpendicular  from  whose  vertex  falls  at  the  middle 
of  the  base.     This  perpendicular  is  called  the  axis. 

503,  A  Frustum  of  a  pyramid  is  a  portion  of  the  pyramid 
intercepted  between  the  base  and  a  plane  parallel  to  the  base.  If 
the  cutting  plane  is  not  pai-allel  to  the  base,  the  portion  intercepted 
is  called  a  Truncated  pyramid. 

504,  Tlie  Slant  Weight  of  a  right  pyramid  is  the  altitude 
of  one  of  the  triangles  which  form  its  faces.  The  Slant  Height  of  a 
Frustum  of  a  right  pyramid  is  the  portion  of  the  slant  height  of  the 
pyramid  intercepted  between  the  bases  of  the  frustum. 


Fig.  303. 


Ill's. — The  student  will  be  able  to  find  iUustrations  of  the  definitions  in  the 
accompanying  figures. 

505.  A  Conical  Surface  is  a  surface  traced  by  a  line  which 

passes  through  a  fixed  point,  while  any  other  point  traces  a  curve. 
The  line  is  the  Generatrix,  and  the  curve  the  Directrix.  The  fixed 
point  is  the  Vertex.  Any  line  of  the  surface  corresponding  to  some 
position  of  the  generatrix  is  called  an  Element  of  the  surface. 

506.  A  Cone  of  Hevolution  is  a  solid  generated  by  the 
revolution  of  a  right  angled  triangle  around  one  of  its  sides,  called 
the  Axis.  The  hypotenuse  describes  the  Convex  Surface  of  the 
cone,  and  corresponds  to  the  generatiix  in  the  preceding  definition. 
The  other  side  of  the  triangle  describes  the  Base.  This  cone  is  right, 
since  the,  perpendicular  (the  axis)  falls  at  the  middle  of  the  base. 
The  Slant  Height  is  the  distance  from  the  vertex  to  the  circumfer- 
ence of  the  base,  and  is  the  same  as  the  hypotenuse  of  the  generating 
triangle. 

507.  The  terms  Frustum  and  Truncated  qiq  applied  to  the  cone 
in  the  same  manner  as  to  the  pyramid. 


OF  PYRAMIDS  AND  CONES. 


201 


S08^  A  pyramid  is  said  to  be  Inscriled  in  a  cone  when  the  base 
of  the  pyramid  is  inscribed  in  the  base  of  the  cone,  and  the  edges  of 
the  pyramid  are  elements  of  the  surface  of  the  cone.  The  two  solids 
have  a  common  vertex  and  a  common  altitude. 

S09.  If  the  generatrix  be  considered  as  an  indefinite  straight 
line  passing  through  a  fixed  point,  the  portions  of  the  line  on  oppo- 
site sides  of  the  point  will  each  describe  a  conical  surface.  These 
two  surfaces,  which  in  general  discussions  are  considered  but  one,  are 
called  Nappes,  The  two  nappes  of  the  same  cone  are  evidently 
alike. 

Ill's. — In  the  figure,  {a)  represents  a  conical  surface  which  has  the  curve 
ACB  for  its  directrix,  and  SA  for  its  generatrix.    The  figures  indicate  the  suc- 


Fio.  304. 


cessive  positions  of  the  point  A,  as  it  passes  around  the  curve,  while  the  point  S 
remains  fixed.  (&)  represents  a  Cone  of  Bewlution,  or  a  right  cone  with  a  cir- 
cular base.  It  may  be  considered  as  generated  in  the  general  way,  or  by  the 
right  angled  triangle  SOA  revolving  about  SO  as  an  axis.  SA  describes  the 
convex  surface,  and  OA  the  base.  The  figure  (c)  represents  the  Frustum  of  a 
cone,  the  portion  above  the  plane  ahc  being  supposed  removed.  Figure  {d)  rep- 
resents the  two  nappes  of  an  oblique  cone. 


v!- 


PROPOSITION  I. 

510.  TJieorein,—Any  section  of  a  pyramid  made  ly  a  plane 
parallel  to  its  hasc  is  a  polygon  similar  to  the  hase. 


202 


ELEMENTARY   SOLID   GEOMETRY. 


Dem.— The  section  abcde  of  the  pyramid  S-ABCDE,  made  by  a  plane  parallel 
to  ABCDE,  is  similar  to  ABCDE. 

Since  AB  and  db  are  intersections  of  two  parallel 
planes  by  a  third  plane,  they  are  parallel  (?).  So 
also  he  is  parallel  to  BC,  cd  to  CD,  etc.  Hence, 
angle  6  =  B,  c  =  C,  etc.  (?),  and  the  polygons  are 
mutually  equiangular.  Again,  aJ  :  AB  : :  S6  :  SB, 
and  5c  :  BC  : :  S5  :  SB  (?).  Hence,  ab  :  be  ::  AB 
■  BC  (?).  In  like  manner,  we  can  show  that  be  : 
cd  ::  BC  :  CD,  etc.  Therefore,  abcde  and  ABCDE 
are  mutually  equiangular,  and  have  their  corre- 
sponding sides  proportional,  and  are  consequently 
similar.    Q.  E.  d. 


PROPOSITION  n. 

Sll*  Tlieorem, — IfUuo  pyramids  of  the  same  altitude  are  cut 
hy  planes  equally  distant  from  and  'parallel  to  their  bases,  the  sections 

are  to  each  other  as  the  bases. 

Dem.— Let  S-ABC  and  S'-A'B'C'D'E'  be 
two  pyramids  of  the  same  altitude,  cut  by 
the  planes  abc  and  a'b'c'd'e',  parallel  to  and 
at  equal  distances  from  their  bases ;  then  is 
abc  :  a'b'c'd'e'  :  :  ABC   :  A'B'C'D'E'. 

For,  conceive  the  bases  in  the  same 
plane.  Let  SP  =  S'P'  be  the  common  alti- 
tude, and  Sp  =  S'p'  the  distances  of  the 
cutting  planes  from  the  vertex.    We  have 

'-ycs- 

Also,  A'B'C'D'E'  :  a'b'c'd'e'  :  :  A'B''  :  a'6''  : :  SP''  :  Sy'  (?).       ^-  y^ 
Whence,  as  SP  =  S'P',  and  Sp  =  S'p'  (?),  we  have  ^  -^ 

abc  :  a'b'c'd'e'  :  :  ABC  :  A'B'C'D'E'  (?).    q.  e.  d. 

S12.  Cor. — If  the  bases  are   equivalent,   the  sections  are  also 
equivalent. 


Fig.  306. 


ABC  :  abc  -.'.  AB 


ab 


SJ'    :  Sp  (?)  I 


PROPOSITION  III. 

513*  Tlieorem. — TJie  area  of  the  lateral  surface  of  a  right 
pyra7nid  is  equal  to  the  perimeter  of  the  base  m.ultijMed  by  one-half 
the  slant  height. 

^.  Dem.— The  faces  of  such  a  pyramid  are  equal  isosceles  triangles  (?),  whose 
common  altitude  is  the  slant  height  of  the  pyramid  (?).     Hence,  the  area  of 


OF  PYRAMIDS  AND   CONES. 


203 


these  triangles  is  the  product  of  one-half  the  slant  height  into  the  sura  of  their 
bases.  But  this  is  the  lateral  surface  of  the  pyramid.  (See  the  third  cut  in 
Fig.  303.) 

514i,  Cor. — The  area  of  the  lateral  surface  of  the 
frustum  of  a  right  pyramid  is  equal  to  the  product 
of  its  slant  height  into  half  the  sum  of  the  perimeters 
of  its  bases. 

The  student  will  be  able  to  give  the  proof.  It  is  based 
upon  {325)  and  definitions.  Fig.  301 


PROPOSITION  IV. 

51S,  Theorem. — TJie  area  of  the  convex  surf  ace  of  a  cone  of 
revolution  {a  right  cone  with  a  circular  base)  is  equal  to  the  product 
of  the  circumference  of  its  base  and  one-half  its  slant  height,  i.  e., 
ttKH',  R  being  the  radius  of  the  base,  and  H'  the  slant  height. 

Dem. — In  the  circle  which  forms  the  base  of  the  cone,  conceive  a  regular 
polygon  inscribed,  as  dbcde.  Joining  the  vertices  of  the 
angles  of  this  polygon  with  the  vertex  of  the  cone,  there 
will  be  constructed  a  right  pyramid  inscribed  in  the  cone. 
Now,  if  the  arcs  subtended  by  the  sides  of  this  polygon  are 
bisected,  and  these  again  bisected,  etc.,  and  at  every  step 
a  right  pyramid  conceived  as  inscribed,  it  will  always 
remain  true  that  the  lateral  surface  of  the  pyramid  is  the 
perimeter  of  its  base  into  half  its  slant  height.  But, 
as  the  number  of  faces  of  the  pyramid  is  increased, 
the  perimeter  of  the  base  approaches  the  circumference 
of  the  base  of  the  cone,  the  slant  height  of  the  pyi-amid 
approaches  the  slant  height  of  the  cone,  and  the  lateral 
surface  of  the  pyramid  approaches  the  convex  surface 
of  the  cone.  Hence,  at  the  limit  we  still  have  the  same  expression  for  the 
area  of  the  convex  surfiice,  that  is,  the  circumference  of  the  base  multiplied  by 
half  the  slant  height.  Finally,  if  R  is  the  radius  of  the  base,  its  circumference 
is  2;rR,  and  H'  being  the  slant  height,  we  have  for  the  area  of  the  convex  sur- 
face 2;rR  x  iH',  or  ttRH'. 

310,  Cor.  1. — 77ie  area  of  the  convex  surface  of  a  cone  is  also 
equal  to  the  product  of  the  slant  height  into  the  circumference  of  the 
circle  parallel  to  the  base,  and  midway  between  the  base  and  vertex. 

This  follows  directly  from  the  flict  that  the  radius  of  the  circle  midway 
between^the  base  and  vertex  is  one-half  the  radius  of  the  base,  i.  e.,  ^R,  whence 
its  circumference  is  ttR.  Now,  ttR  x  H'  is  the  area  of  the  convex  surface,  by 
the  proposition. 


204 


ELEMENTARY  SOLID  GEOMETRY. 


517,  Cor.  2. — Tlie  area  of  the  convex  surface  of  tlie  frustum,  of  a 
cone  is  equal  to  the  product  of  its  slant  height  into  half  the  sum  of 
the  circumferences  of  its  bases;  i.  e.,  ?r  (R  +  r)  H',  R  and  r  being 
the  radii  of  its  bases,  and  H'  its  slant  height. 

Pmm  the  corresponding  property  of  the  fi-ustum  of  a  pyramid,  the  student 
will  be  able  to  deduce  the  fact  that  i  (2;rR  +  2itr)  H',  or  -n-  (R  +  r)  H',  is  the 
area  of  this  surface. 

518,  Cor.  3. — The  area  of  the  convex  surface  of  the  frustum  of  a 
cone  is  equal  to  the  product  of  its  slant  height  into  the  circumference 
of  the  circle  midway  between  the  bases. 

The  radius  of  the  circle  midway  between  the  bases  is  ^  (r  +  R),  whence  its 
circumference  is  n  {r  +  R).  Now,  tt  (r  +  R)  x  H'  is  the  area  of  the  convex 
surface  of  the  frustum,  by  the  preceding  corollary. 


PROPOSITION  V. 


519.  TJieorem, — Two  pyramids  having  equivalent  bases  and 
the  same  altitudes  are  equivalent,  i.  e.,  equal  in  volume. 

Dem.— Let  S-ABCD  and  S'-A'B'C'D'E'  be  two  pyramids  having  the  same 

altitudes,  and  base  ABCD  equivalent 
to  base  A'B'C'D'E',  i.  e.,  equal  in  area; 
then  is  pyramid  S-ABCD  equivalent 
to  S'-A'B'C'D'E',  i.  e.,  equal  in  volume. 
For,  conceive  the  bases  to  be  in  the 
same  plane,  and  a  plane  to  start  from 
coincidence  with  the  plane  of  the 
bases,  and  move  toward  the  vertices, 
remaining  all  the  time  parallel  to  the 
bases.  At  every  stage  of  its  progress 
the  sections  are  equivalent,  and  as  the 
plane  reaches  both  vertices  at  the 
same  time,  by  reason  of  the  common  altitude,  it  is  evident  that  the  volumes  are 
equal. 

Or,  if  desired,  we  may  consider  the  two  pyramids  as  divided  into  an  equal 
number  of  infinitely  thin  lam^.ncB  parallel  to  the  bases.  Each  lamina  in  one  l^as 
its  corresponding  equivalen*^  lamina  in  the  other ;  hence  the  sum  of  all  the 
lamince  in  one  equals  the  sum  of  all  the  lamince  in  the  other ;  i.  «.,  the  pyramids 
are  equivalent. 


OP  PYRAMIDS  AND  CONES, 


205 


PROPOSITION  VI. 


520.  Uieorem. — TJie  volume  of  a  triangular  pyramid  is  equal 
to  one-third  the  product  of  its  base  and  altitude. 


Dem. — Let  S-ABC  be  a  triangular  pyramid, 
whose  altitude  is  H* ;  then  is  the  volume  equal 
to  ^  H  X  area  ABC. 

For,  throui^h  A  and  B  draw  Aa  and  Bh  paral- 
lel to  SC;  and  through  S  draw  Sa  and  Sh 
parallel  to  CA  and  CB,  and  join  a  and  h\  then 
Sdb-kBC  is  a  prism  with  its  bases  equal  to  the 
base  of  the  pyramid.  Now,  the  solid  added  to 
the  given  pyramid  is  a  quadrangular  pyramid 
with  obBK  as  its  base,  and  its  vertex  at  S.  Divide 
this  into  two  triangular  pyramids  by  draw- 
ing aB  and  passing  a  plane  through  SB  and 
aB.  These  triangular  pyramids  are  equiva- 
lent, since  they  have  equal  bases  aAB  and  abB 
and  a  common  altitude,  the  vertices  of  both  being  at  S 


Fio.  310. 


Again,  S-ahB  may  be 
considered  as  having  abS  (equal  to  ABC)  as  its  base,  and  the  altitude  of  the  first 
pyramid  (equal  to  the  altitude  of  the  prism)  for  its  altitude,  and  hence  as 
equivalent  to  the  given  pyramid.  Therefore  S-ABC  is  one  third  of  the  prism 
Sdb-kBC.  But  the  volume  of  the  prism  is  H  x  area  ABC.  Therefore  the 
volume  of  the  pyramid  S-ABC  is  ^  H  x  area  ABC.    q.  e.  d. 

521.  Cor.  1. — The  volume  of  any  pyramid  is  equal  to  one-third 
the  product  of  its  base  and  altitude. 

Dem. — Since  any  pyramid  can  be  divided  into  trian- 
gular pyramids  by  passing  planes  through  any  one  edge, 
as  SE,  and  each  of  the  other  edges  not  adjacent,  as  SB  and 
SC,  the  volume  of  the  pyramid  is  equal  to  the  sum  of  the 
volumes  of  several  triangular  pyramids  having  the  same 
altitude  as  the  given  pyramid,  and  the  sum  of  whose  bases 
is  the  base  of.  the  given  pyramid.  Hence  the  truth  of  the 
corollaiy. 

322,  Cor.  2. — Pyra^nids  having  equivalent  bases 
are  to  each  other  as  their  altitudes;  such  as  have  equal  altitudes 
are  to  each  other  as  their  bases;  and,  in  general,  jjyramids  are 
to  each  other  as  the  products  of  their  bases  and  altitudes. 


*  Not  drawn  in  the  figure,  leet  it  might  confuae. 


206 


ELEMENTARY   SOLID  GEOMETRY. 


Fi6.  312 


PROPOSITION  TIL 

523,  Tlieorein, —  Tlie  volume  of  the  frustum  of  a  triangular 
pyramid  is  equal  to  the  volume  of  three  pyramids  of  the  same 
altitude  as  the  frustum,  and  lohose  bases  are  the  upper  base,  the  lower 
base,  and  a  mean  proportional  betioeen  the  two  bases  of  the  frustum, 

Dem. — Let  a6c-ABC  be  the  fnistiim  of  a  triangu- 
lar pyramid.  Through  a6  and  C  pass  a  plane  cutting 
off  the  pyramid  C-abc.  This  has  for  its  base  the 
upper  base  of  the  frustum,  and  for  its  altitude  the 
altitude  of  the  fnistum.  Again,  draw  A6,  and  pass 
a  plane  through  A6  and  bC,  cutting  off  the  pyramid 
6-ABC,  -which  has  the  same  altitude  as  the  frustum, 
and  for  its  base  the  lower  base  of  the  frustum. 
There  now  remains  a  third  pyramid,  b-ACa,  to  be  ex- 
amined. Through  b  draw  bD  parallel  to  aA,  and 
draw  DC  and  aD.  The  pyramid  D-ACa  is  equiva- 
lent to  6-ACa,  since  it  has  the  same  base  and  the 
same  altitude.  But  the  former  may  be  considered  as 

Jiaving  ADC  for  its  base,  and  the  altitude  of  the  fnistum  for  its  altitude,  i.  e., 

as  pyramid  a-ADC.    We  are  now  to  show  that  ADC  is  a  mean  proportional 

between  abc  and  ABC. 

ABC   :  abc  :  : 

Also,  ABC  :  ADC 

whence  ABC*  :  ADC' 

By  equality  of  ratios,  ABC  :  abc  ::  ABC'  :  ADC*; 

whence  ADC''  =  abc  x  ABC,  i.  e.,  ADC  is  a  mean  proportional  between  the 

upper  and  lower  bases  of  the  frustum. 

524.  Cor. — The  volume  of  the  frustum  of  any  pyramid  is 
equal  to  the  volume  of  three  pyramids  having  the  same  altitude  as 
the  frustum,  and  for  bases,  the  upper  base,  the  loioer  base,  and  a 
mean  proportional  between  the  bases  of  the  frustum. 

For,  the  frustum  of  any  pyramid  is  equivalent  to  the  corresponding  frustum 
of  a  triangular  pyramid  of  the  same  altitude  and  an  equivalent  base  (?) ;  and 
the  bases  of  the  frustum  of  the  triangular  pyramid  being  both  equivalent  to 
the  corresponding  bases  of  the  given  frustum,  a  mean  proportional  between 
the  triangular  bases  is  a  mean  proportional  between  their  equivalents. 


AB' 
AB 


ab    :  :  AB    :  AD    (?). 
AD  (?); 


AB    :  AD    (?). 


PROPOSITION  vm. 

S2S.  TJieorem, — The  volume  of  a  cone  of  revolution  is  equal  to 
one-third  the  product  of  its  base  and  altitude  ;  i.  e.,  ^;rR'H,  R  being 
the  radius  of  the  base  and  H  the  altitude. 


OF  PYRAMIDS  AND  CONES.  207 

Dem. — This  follows  from  the  volume  of  a  pyramid,  by  a  course  of  reasoning 
precisely  the  same  as  in  {515).  The  volume  of  a  pyramid  being  equal  to  one- 
third  the  product  of  the  base  and  altitude,  and  the  cone  being  the  limit  of 
the  pyramid,  the  volume  of  the  cone  is  one-third  the  product  of  its  base  and 
altitude.  -  Now,  R  being  the  radius  of  the  base  of  a  cone  of  revolution,  the 
base  (area  of)  is  ttR",  whence  i^rR^H  is  the  volunae,  H  being  the  altitude. 

526 »  Cor.  1. — The  volume  of  any  cone  is  equal  to  one-third  the 
prodilct  of  its  base  and  altitude. 

S27,  Cor.  2. — The  volume  of  the  frustum  of  a  cone  is  equal  to 
the  volume  of  three  cones  having  the  same  altittide  as  the  frustum, 
and  for  bases,  the  tipjjer  base,  the  lower  base,  and  a  mean  propor- 
tional between  the  two  bases  of  the  frustum. 

The  truth  of  this  appears  from  the  fact  that  the  frustum  of  a  cone  is  the 
limit  of  the  frustum  of  a  pyramid. 


PROPOSITION  IX. 

S28,  TJieorein, — The  lateral  surfaces  of  similar  right  pyra- 
mids are  to  each  other  as  the  squares  of  their  homologous  edges,  their 
slant  heights,  and  their  altitudes  ;  i.  e.,  as  the  squares  of  any  two 
homologous  dimensions. 

Dem. — Let  A  and  a  be  homologous  sides  of  the  bases  of  two  similar  right 
pyramids,  H'  and  h'  their  slant  heights,  H  and  h  their  altitudes,  and  P  and  p 
tJie  perimeters  of  their  bases ;  then — 

(1)  P  :  _p  : :  A  :  a,  because  the  bases  are  similar  polygons ; 

(2)  A  :  a  : :  H'  :  h\  because  the  faces  are  similar  triangles ; 

(3)  H'  :  A'  :  :  H  :  h  (?). 
(Vhence,  P  :  p  :  :  H'  :  A' ; 
and,  as            ^H'  :^h' ::  H' :  h', 

multiplying,   we  have  ^P  x  H' :  \px  h'  :  :  H'^  h"" :  :  A'^ :  a^ : :  H' :  li".     But 
tP X  H'  and  ^pxN  are  the  areas  of  the  lateral  surfaces. 


PROPOSITION  X. 

529*  TJieoretn, — The  convex  surfaces  of  similar  cones  of  revo- 
lution are  to  each  other  as  the  sqicares  of  their  slant  heights,  the  radii 
of  their  bases,  and  their  altitudes  ;  i.  e.,  as  the  squares  of  any  two  ho- 
mologous dimensions,  '^ 

Dem.— Let  H'  and  h'  be  the  slant  heights  of  two  similar  cones  of  revolution, 
R  and  r  the  radii  of  their  bases,  and  H  and  h  their  altitudes;  their  convex 
surfaces  are  ;rRH'  and  Ttrh'.    Now,  since  the  cones  are  similar  R  :  r  : :  H'  :  /t'. 


208  ELEMENTARY  SOLID   GEOMETRY. 

Multiplying  the  terms  of  this  proportion  by  the  corresponding  terms  of  nW  • 
ich'  : :  H'  :  h\  we  have — 

;rRH'  :  TtrJi'  : :  H"  :  7t'\ 
Hence  the  convex  surfaces  are  as  the  squares  of  the  slant  heights,  and  since 
B:  r  ::  R'  '.  h'  ::  H   :  h  (?),  R»  :  r»  :  :  H*  ;  h""  : :  H^  :  h' ;   and  consequently 
;rRH'  :  xrh'  :  :  R^  :  r'  : :  H^  :  h\ 


PROPOSITION  XL     Y?^      ^ 


S30»  TJieorem. — The  volumes  of  similar  pyramids  are  to  each 
other  as  the  cubes  of  their  homologous  dimensions. 

Dem. — Letting  A  and  a  be  homologous  sides  of  the  bases  of  two  similar 
pyramids,  B  and  b  their  bases,  and  H  and  h  their  altitudes,  the  student  should 
be  able  to  give  the  reasons  for  the  following  proportions  : 

^H  :  ^A  :  :  A  :  a  : :  H  :  7^. 
Whence  iBH  :  \hh  : :  A.^  :  a^  : -.  W  :  K\     Q.  E.  D. 


PROPOSITION  xn.     j^    ^ 

531,  Tlieoretn, — Tlie  volumes  of  similar  cones  are  to  each  other 
as  the  cubes  of  their  altitudes,  or  as  the  cubes  of  the  radii  of  their 
bases, 

Dem.    R  and  r  being  the  radii  of  their  bases,  and  H  and  h  their  altitudes, 
R»  ^:  r*  ::  ff  :  A»  (?),  and  R"  :  r^  : :  H^*  :  A^ 
Also,  ^,7rR:^7ih::R:h. 

Multiplying,  ^tzR-'R  :  ^itr'h  :  :  H'  :  h\  or  as  R'  :  J'\     Q.  e.  d. 


EXERCISES. 

1.  What  is  the  area  of  the  lateral  surface  of  a  right  hexagonal 
pyramid  whose  base  is  inscribed  in  a  circle  whose  diameter  is  20  feet, 
the  altitude  of  the  pyramid  being  8  feet  ?  What  is  the  volume  of 
this  pyramid  ? 

2.  What  is  the  area  of  the  lateral  surface  of  a  right  pentagonal 
pyramid  whose  base  is  inscribed  in  a  circle  whose  radius  is  6  yards, 
the  slant  height  of  the  pyramid  being  10  yards  ?  What  is  the  vol- 
ume of  this  pyramid  ? 

3.  How  many  quarts  will  a  can  contain,  whose  entire  height  is  10 
inches,  the  body  being  a  cylinder  6  inches  in  diameter  and  6 J  inches 


OF   THE   SPHERE.  209 

high,  and  the  top  a  cone  ?   How  much  tin  does  it  take  to  make  such 
a  can,  allowing  nothing  for  waste  and  the  seams  ? 

4.  If  very  fine  dry  sand  is  piled  upon  a  smooth  horizontal  surface,  / 
without  any  lateral  support,  the  angle  of  slope  {i.  e.,  the  angle  of 
inclination  of  the  sloping  side  of  the  pile  with  the  plane)  is  about  31°. 
Suppose  two  circles  be  drawn  on  the  floor,  one  4  feet  in  diameter  and 
the  other  3,  and  sand  piles  be  made  as  large  as  possible  on  these  cir- 
cles as  bases,  no  other  support  being  given.  What  is  the  relative 
magnitude  of  the  piles  ? 

5.  In  the  case  of  sand  piles,  as  given  in  the  last  example,  the  ratio 
of  the  radius  of  the  base  to  the  altitude  of  the  pile  is  f.  How  many 
cubic  feet  in  each  of  the  above  piles  ? 

6.  The  frustum  of  a  right  pyramid  was  72  feet  square  at  the  lower 
base  and  48  at  the  upper ;  and  its  altitude  was  60  feet.  What  was 
the  lateral  surface  ?    What  the  volume  ? 


The  student  should  furnish  a  synopsis  of  each  section  at- 

ITS  close. 


SECTION  V. 

OF    THE    SPHERE* 


532,  A  Sphere  is  a  solid  bounded  by  a  surface  every  point  in 
which  is  equally  distant  from  a  point  within  called  the  Centre.  The 
distance  from  the  centre  to  the  surface  is  the  Radius,  and  a  line 
passing  through  the  centre  and  limited  by  the  surface  is  a  Diameter. 
The  diameter  is  equal  to  twice  the  radius. 


*  a  epherical  blackboard  is  almost  indispensable  in  teaching  this  section,  as  well  as  in 
teaching  Spherical  Trigonometry.  A  sphere  about  2  feet  in  diameter,  mounted  on  a  pedestal, 
and  having  its  surface  elated  or  painted  as  a  blackboard,  is  what  is  needed.  It  can  be  ob« 
tained  of  the  manufacturers  of  echool  apparatus,  or  made  In  any  good  tumiug-shop. 


210 


ELEMENTARY  SOLID  GEOMETRY. 


CIRCLES  OF  THE  SPHERE. 


PROPOSITION  L 

533.  Tlieorem. — Every  section  of  a  sphere,  made  hy  a  plane,  ti 
a  circle. 

Dem.— Let  AFEBD  be  a  section  of  a  sphere 
•whose  centre  is  0,  made  by  a  plane ;  then  is  it  a 
circle. 

For,  let  fall  from  the  centre  O  a  pei-pendicular 
upon  the  plane  AFEBD,  as  OC,  and  draw  CA,  CD, 
CE,  CB,  etc.,  lines  of  the  plane,  from  the  foot  of  the 
perpendicular  to  any  points  in  which  the  plane 
cuts  the  surface  of  the  sphere.  Join  these  points 
with  the  centre,  O,  of  the  sphere.  Now,  OA,  OD, 
OB,  OE,  etc.,  beini^  radii,  are  equal;  whence,  CA, 
CD,  CB,  CE,  etc.,  are  equal ;  i. «.,  every  point  in  the 
line  of  intei-section  of  a  plane  and  surface  of  a 

sphere  is  equally  distant  from  a  point  in  this  plane.    Hence,  the  intersection  is 

a  circle,    q.  e.  d. 

534.  Def. — A  circle  made  by  a  plane  not  passing  tli rough  the 
centre  is  a  Small  Circle  ;  one  made  by  a  plane  passing  through  the 
centre  is  a  Great  Circle. 

535.  Cor.  1. — A  perpendicular  from  the  centre  of  a  sphere,  upon 
any  small  circle,  pierces  the  circle  at  its  centre ;  and,  conversely,  a 
perpendicular  to  a  small  circle  at  its  centre  passes  through  the  cefitre 
of  the  sphere. 

536.  Def. — A  diameter  perpendicular  to  any  circle  of  a  sphere 
is  called  the  Axis  of  that  circle.  The  extremities  of  the  axis  are 
the  Poles  of  the  circle. 

537.  Cor.  2. — The  pole  of  a  circle  is  equally  distant  from  every 
point  in  its  circumference. 

The  student  should  be  able  to  give  the  reason. 

538.  Cor.  3. — Every  circle  of  a  sphere  has  two  poles,  lohich,  in 
case  of  a  great  circle,  are  equally  distant  from  every  point  in  the  cir- 
cumference of  the  circle ;  but,  in  case  of  a  small  circle,  one  pole  is 
nearer  any  poiiit  in  the  circumference  tha^i  the  other  pole  is. 


DISTANCES   ON  THE   SURFACE  OP  A  SPHERE.  211 

SS9,  Cor.  4. — A  small  circle  is  less  as  its  distance  from  the  cen- 
tre of  the  sphere  is  greater. 

For, its  diameter,  being  a  chord  of  a  great  circle,  is  less  as  it  is  farther  fi'om 
the  centre  of  the  great  circle,  which  is  also  the  centre  of  the  sphere. 

S4:0.  Cor.  5. — All  great  circles  of  the  same  sphere  are  equal,  their 
radii  being  the  radius  of  the  sphere. 


PROPOSITION   n. 

541.  Theorem, — Any  great  circle  -divides  the  sphere  into  two 
equal  parts  called  Hemispheres. 

Dem. — Conceive  a  sphere  as  divided  by  a  great  circle,  i.  «.,  by  a  plane  passing 
through  its  centre,  and  let  the  great  circle  be  considered  as  the  base  of  each 
portion.  These  bases  being  equal,  reverse  one  of  the  portions  and  conceive 
its  base  placed  in  the  base  of  the  other,  the  convex  surfaces  being  on  the  same 
side  of  the  common  base.  Since  the  bases  are  equal  circles,  they  will  coincide, 
and  since  every  point  in  the  convex  surface  of  each  portion  is  equally  distant 
from  the  centre  of  the  common  base,  the  convex  surfaces  will  coincide.  There- 
fore, the  portions  coincide  throughout,  and  ai-e  consequently  equal,    q.  e.  d. 


PROPOSITION  m. 

542.  Ttieorem. — TJie  intersection  of  any  two  great  circles  of  a 
sphere  is  a  diameter  of  the  sphere. 

Dem. — The  intersection  of  two  planes  is  a  straight  line ;  and  in  the  case  of 
the  two  great  circles,  as  they  both  pass  through  the  centre  of  the  sphere,  this  is 
one  point  of  their  intersection.  Hence,  the  intersection  of  two  great  circles  of 
a  sphere  is  a  straight  line  which  passes  through  the  centre.    Q.  e.  d. 

543.  Cor. — The  intersections  on  the  surface  of  a  sphere  of  two 
circumferences  of  great  circles  are  a  semi-circumference,  or  180°, 
cepart,  sifice  they  are  at  opposite  extremities  of  a  diameter. 


DISTANCES  ON  THE  SURFACE  OF  A  SPHERE. 

544.  Distances  on  the  surface  of  a  sphere  are  always  to  be  under- 
stood as  measured  on  the  arc  of  a  great  circle,  unless  it  is  otherwise 
stated. 


212 


ELEMENTARY  SOLID  GEOMETRY. 


(S 


PROPOSITION  IT. 

S4S,  TJieorem, — The  distances,  measured  on  the  surface  of  a 
sphere,  from  a  pole  to  all  p>oints  in  the  circumference  of  a  circle  of 

which  it  is  the  pole,  are  equal. 

Dem.— Let  P  be  a  pole  of  the  small  circle  AEB  ; 
then  are  the  arcs  PA,  PE,  PB,  etc.,  which  measure 
the  distances  on  the  surface  of  the  sphere,  from  P 
to  any  points  in  the  circumference  of  circle  AEB, 
equal.  For,  by  (537),  the  straight  lines  AP,  PE, 
PB,  etc.,  are  eqnaA,  and  these  equal  chords  subtend 
equitl  arcs,  as  arc  PA,  arc  PE,  arc  PB,  etc.,  the  great 
circles  of  which  these  lines  are  chords  and  arcs 
being  equal  {540).  Thus,  for  like  reasons,  arc 
P'QA  =  arc  P'LE  =  arc  P'RB,  etc. 

54:6.  Cor. — TJie  distance  from  the  pole  of  a  great  circle  to  any 
point  in  the  circumference  of  the  circle  is  a  quadrant  {a  quarter  of  a 
circumference). 

Since  the  poles  are  180°  apart  (being  the  extremities  of  a  diameter),  PAQP'  = 
PELF'  =  a  semicircumference.  But,  in  case  of  a  great  circle,  chord  PL  =  chord 
.P'L  (=  chord  PQ  —  chord  P'Q),  whence  arc  PEL  =  arc  P'L  =  arc  PAQ  =  arc 
PQ.     Hence,  each  of  these  arcs  is  a  quadrant. 

547 •  ScH. — By  means  of  the  facts  demonstrated  in 
this  proposition  and  corollary,  we  are  enabled  to  draw 
arcs  of  small  and  great  circles,  in  the  surface  of  a  sphere, 
with  nearly  the  same  facility  as  we  draw  arcs  and 
lines  in  a  plane.  Thus,  to  draw  the  small  circle  AEB, 
we  take  an  arc  equal  to  PE,  and  placing  one  end  of  it 
at  P,  c^use  a  pencil  held  at  the  other  end  to  ti*ace  the 
arc  AEB,  etc.  To  describe  the  circumference  of  a  great 
circle,  a  quadrant  must  be  used  for  the  arc.  By  bend- 
ing a  wire  into  an  arc  of  the  circle,  and  making  a  loop 

in  each  end,  a  wooden  pin  can  be  put  through  one  loop  and  a  crayon  through 

the  other,  and  an  arc  drawn  as  represented  in  the  figure. 


Fig.  315. 


PROPOSITION  T. 

548.   JProhlem, — To  pass  a  circumference  of  a  great  circU 
through  any  two  points  on  the  surface  of  a  sphere. 


DISTANCES  ON  THE  SURFACE  OF  A  SPHERE.  213 

Solution". — Let  A  and  B  be  two  points  on  the  sur- 
face of  a  sphere,  through  which  it  is  proposed  to  pass  a 
circumference  of  a  great  circle.  From  B  as  a  pole,  with_ 
an  arc  ^gual  to  a  quadrant,  strike  an  arc  on^  as  nearly 
wliere  the  pole  of  tne  circle  passing  through  A  and  B 
lies,  as  may  be  determined  by  inspection.  Then,  from 
A,  with  the  same  arc,  strike  an  arc  %t  intersecting  on  at 
P.  NowJPjs  J,he  pole  of  the  great  circlej^assing^^hrmj^h^ 
A  and  B.  Hence,  from  P  as  a  pole,  with  a  quadrant  arc 
*"ttfSw  a  circle;  itwiJiJL  pass  through  A  and  B,  and  will  Fig.  316. 

be  a  great  circle,  since  its  i)ole  is  a  quadrant's  distance 

frOnTTtrcircumference.     [The  student  should  make  the  construction  on  the 
spherical  blackboard.] 

*5^4i>.  Cor.  1. — Throuyh  any  two  points  on  the  surface  of  a  sphere, 
one  great  circle*  can  always  be  made  to  pass,  and  only  one,  except 
luhen  the  two  points  are  at  the  extremities  of  the  same  diameter,  in 
tohich  case  an  infinite  nuniber  of  great  circles  can  le passed  through 
the  tivo  points. 

Since  the  arcs  on  and  st  are  arcs  of  great  circles,  the  circumferences  of  which 
they  form  parts  will  intersect  also  on  tlie  opposite  side  of  the  sphere,  at  a  dis- 
tance of  a  semicircumference  from  P.  But  these  two  points  are  poles  of  the 
same  great  circle.  Now,  as  the  two  great  circles  can  intersect  at  no  other  points, 
there  can  be  only  one  great  circle  passed  through  A  and  B.  But  if  the  two 
given  points  were  at  the  extremities  of  the  same  diameter,  as  at  D  and  C,  the 
arcs  s^and  on  would  coincide,  and  any  point  in  this  circumference  being  taken 
as  a  pole,  great  circles  can  be  drawn  through  D  and  C.  [The  student  should 
ti'ace  the  work  on  the  spherical  blackboard.] 

550,  SCH. — The  truth  of  the  corollary  is  also  evident  from  the  fact  that 
three  points  not  in  the  same  straight  line  determine  the  position  of  a  plane. 
Thus  A,  B,  and  tlie  centre  of  the  sphere,  fix  the  position  of  one,  and  only  one, 
great  circle  passing  through  A  and  B.  Moreover,  if  the  two  given  points  are  at 
the  extremities  of  the  same  diameter,  they  are  in  the  same  straight  line 
with  the  centre  of  the  sphere,  whence  an  infinite  number  of  planes  can  be 
passed  through  them  and  the  centre.  The  meridians  on  the  earth's  surface  af- 
ford an  example,  the  poles  (of  the  equator)  being  the  given  points. 

551.  Cor.  2. — If  two  points  in  the  circumference  of  a  great  circle 
of  a  sphere,  not  at  the  extremities  of  the  same  diameter,  are  at  a 
quadranfs  distance  from  a  point  on  the  surface,  that  point  is  the 
pole  of  the  circle. 

*  The  word  circle  may  be  understood  to  refer  either  to  the  circle  proper,  or  to  its  cir- 
cumference.   The  word  is  in  constant  use  in  the  higher  mathematics,  in  the  latter  sense. 


214 


ELEMENTARY  SOLID   GEOMETRY. 


PROPOSITION  Tl. 


Fig.  31'; 


552.  Theorem, — TJie  shortest  distance  on  the  surface  of  a 
sphere,  hetiveen  any  two  points  in  that  surface,  is  measured  on  the  arc 
less  than  a  s&inicircumference  of  the  great  circle  which  joins  them. 

Dem.— Let  A  and  B  be  any  two  points  in  the  sur- 
face of  a  sphere,  AB  the  arc  of  a  great  circle  joining 
them,  and  AmC^tB  any  other  path  in  the  surface  be- 
tween A  and  B ;  then  is  arc  AB  less  than  kmCiiB. 

Let  C  be  any  point  in  A7nCnB,  and  pass  the  arcs  of 
great  circles  through  A  and  C,  and  B  and  C.  Join  A, 
B,  and  C  with  the  centre  of  the  sphere.  The  angles 
AOB,  AOC,  and  COB  form  the  facial  angles  of  a  trie- 
dral,  of  which  angles  the  arcs  AB,  AC,  and  CB  are  the 
measures.  Now,  angle  AOB  <  AOC  +  COB  (454); 
whence  arc  AB  <  arc  AC  +  arc  CB,  and  the  path  from 
A  to  B  is  less  on  arc  AB  than  on  arcs  AC,  CB.  In  like  manner,  joining  any  point 
in  iKmC  with  A  and  C  by  arcs  of  great  circles,  their  sum  would  be  greater  than 
AC.  So,  also,  joining  any  point  in  C/iB  with  C  and  B,  the  sum  of  the  arcA 
would  be  greater  than  CB.  As  this  process  is  indefinitely  repeated,  the  path 
from  A  to  B  on  the  arcs  of  the  great  circles  will  continually  increase,  and  also 
continually  approximate  the  path  kmCnB.  Hence,  arc  AB  is  less  than  the 
path  A/wCtiB.    q.  e.  d. 

553.  Cor. — Hie  least  arc  of  a  circle  of  a  sphere  joining  any 

tivo  points  in  the  surface,  is  the  arc  less  than 
a  semicircumference  of  the  great  circle  pass- 
ing through  the  points  ;  and  the  greatest  arc 
is  the  circumference  minus  this  least  arc. 

Dem.— Let  kniBn  be  any  small  circle  passmg 
through  A  and  B,  and  ABDoC  the  great  circle.  As 
shown  above,  A^^B  <  kmS.  Now,  circumference 
ABDoC  >  circumference  kmBn{539).  Subtracting 
the  former  inequality  from  the  latter,  we  have 
BD<>CA  >  BnA.    q.  e.  d. 


Fig.  318. 


PROPOSITION  vn. 

554.  T1ieorein.—The  shortest  path  on  the  surface  of  a  hemi- 
sphere, from  any  point  therein  to  the  circumfereiice  of  the  great  circle 
f  mining  its  base,  is  the  arc  less  than  a  quadrant  of  a  great  circle  per- 
pendicular to  the  base,  and  the  longest  path,  on  any  arc  of  a  great 
circle,  is  the  supplement  of  this  shortest  path. 


SPHERICAL  ANGLES. 


215 


Fig.  319. 


Dem. — Let  P  be  a  point  in  the  surface  of  the  hemi- 
sphere whose  base  is  ACBC,  and  DPmD'  an  arc  of  a 
great  circle  passing  through  P  and  perpendicular  to 
ADCBC ;  then  is  PD  the  shortest  path  on  the  surface 
from  Pto  circumference  ADBC,  and  PmD'  is  the 
longest  path  from  P  to  the  circumference,  measured 
on  the  arc  of  a  great  circle. 

For,  the  shortest  path  from  P  to  any  point  in  cir- 
cumference ADBC  is  measured  on  the  arc  of  a  great 
circle  {552).  Now,  let  PC  be  any  oblique  arc  of  a 
great  circle.  We  will  show  that  arc  PD  <  arc  PC.  Pro- 
duce PD  until  DP'  —  PD  ;  and  pass  a  great  circle  through  P'  and  C.  Draw  the 
radii  OP,  OD,  OC,  and  OP'.  The  triedrals  0-PDC  and  0-P'DC  have  the  facial  angle 
POD  =  P'OD,  they  being  measured  by  equal  arcs,  and  the  facial  angle  DOC  com- 
mon. Hence,  as  the  included  diedrals  are  equal,  both  being  right,  the  triedrals  are 
equal  or  symmetrical  (440).  In  this  case  they  are  symmetrical,  and  the  facial 
angle  POC  =  P'OC  ;  whence  the  arc  PC  =  arc  P'C.  Finally,  since  PC  -i-  P'C  > 
PP',  PC,  the  half  of  PC  +  P'C,  is  greater  than  PD,  the  half  of  PP'. 

Secondly,  PinD'  is  the  supplement  of  PD,  and  we  are  to  show  that  it  is  greater 
than  any  other  arc  of  a  great  circle  from  P  to  the  circumference  ADBC.  Let 
PtiC  be  any  arc  of  a  great  circle  oblique  to  ADCBC.  Produce  CwP  to  C.  Now 
CPiiC  is  a  seniicircumference  and  consequently  equal  to  DPmD'.  But  we  have 
before  shown  that  PD  <  PC,  and  subtracting  these  from  the  equals  CP?iC  and 
DPmD',  we  have  PmD'  >  P;iC'. 

SS5.  Cor. — Fro77i  any  point  m  the  surface  of  a  liemispliere  there 
are  two  pe7pe7idiculars  to  the  circumference  of  the  great  circle  ^vhich 
forms  the  base  of  the  hemisphere ;  one  of  which  i^erpendiculars 
measures  the  least  distance  to  that  circumference,  and  the  other  the 
greatest,  on  the  arc  of  any  great  cii'cle  of  the  S2)here. 

Thus  PD  and  PmD'  are  two  perpendiculars  from  P  upon  the  circumference 
ADBC. 


SPHERICAL  ANGLES. 

S36,  The  angle  formed  by  two  arcs  of_ 
Jfcircles  of  a  sphere  is  conceived  as  the  same 
as  the  angle  included  by  the  tangents  to 
the  arcs  at  the  common  point. 

III. — Let  AB  and  AC  be  two  arcs  of  circles  of 
the  sphere,  meeting  at  A  ;  then  the  angle  BAC  is 
conceived  as  the  same  as  the  angle  B'AC,  B'A 
being  tangent  to  the  circle  BADm,  and  CA  to  the 
circle  CAEn,. 


Fig.  330. 


216 


ELEMENTARY  SOLED  GEOMETRY. 


Fig.  321. 


557.  A  Splierical  Angle  is  the 

angle  included  by  two  arcs  of  great  circles. 

III. — BAC,  Fig.  321,  is  a  spherical  angle,  and  is 
conceived  as  tlie  same  as  the  angle  B'AC,  B'Aand 
C'A  being  tangents  to  ihe  great  circles  BADF  and 
CAEF.  [The  student  should  not  confound  such  an 
angle  as  BAC,  Fig.  320,  with  a  spherical  angle.} 


Fig.  3^2. 


PROPOSITION  Tin. 

558,  Uieorem, — A  spherical  angle  is  equal  to  the  measure  of 
the  diedral  included  hy  the  great  circles  whose  arcs  form  the  sides  of 
the  angle. 

Dem. — Let  BAC  be  any  spherical  angle,  and 
BADF  and  CAEF  the  great  circles  whose  arcs  BA 
and  CA  include  the  angle  ;  then  is  BAC  equal  to 
the  measure  of  the  diedral  C-AF-B.  For,  since  two 
great  circles  intersect  in  a  diameter  {542),  AF  is 
a  diameter.  Now  B'A  is  a  tangent  to  the  circle 
BADF,  that  is,  it  Hes  in  the  same  plane  and  is  per- 
pendicular to  AG  at  A.  In  like  manner  C'A  lies 
in  the  plane  CAEF  and  is  perpendicular  to  AG. 
Hence  B'AC  is  the  measure  of  the  diedral  C-AF-B 
{4:25).  Therefore  the  spherical  angle  BAC,  which  is  the  same  as  the  plane  angle 
B'AC,  is  equal  to  the  measure  of  the  diedral  C-AF-B-     q.  e.  d. 

559.  Cor.  1. — If  one  of  two  great  circles  passes  through  the  pole 
of  the  other,  their  circumferences  intersect  at  right  angles. 

Dem. — Thus,  P  being  the  pole  of  the  great  circle 
CABm,  PO  is  its  axis,  and  any  plane  passing  through 
PG  is  perpendicular  to  the  plane  CAB;n  {427). 
Hence,  the  diedral  B-AG-P  is  right,  and  the  spheri- 
cal angle  PAB,  which  is  equal  to  the  measure  of  the 
diedral,  is  also  right. 

560*  Cor.  2. — A  spherical  angle  is  meas- 
ured by  the  arc  ofja  great  circle  intercepted 
heticeen  its  sides,  and  at  a  quadrayifs  dis- 
tance from  its  vertex. 

Thus,  the  spherical  angle  CPA  is  measured  by  CA,  PC  and  PA  being  quad- 
rants. For,  since  PC  is  a  quadrant,  CG  is  perpendicular  to  PG,  the  edge  of  the 
diedral  C-PG-A,  and  for  a  like  reason  AG  is  perpendicular  to  PG.  Hence,  CGA 
is  the  measure  of  the  diedral,  and  consequently  CA,  its  measure,  is  the  measure 
of  the  spherical  angle  CPA. 


SPHERICAL  ANGLES. 


\l 


217 


S61*  Cor.  3. — The  angle  included  by  two  arcs  of  ^nqjl  circles  is 
I  the  same  as  the  angle  included  by  two  arcs  of  great  circles  passing 
\i  through  the  vertex  and  having  the  same  tangents. 

Thus  BAC  =  B"AC".  For  the  angle  BAC  is,  by 
definition,  the  same  as  B'AC,  B'A  and  C'A  being 
tangents  to  BA  and  CA.  Now,  passing  planes 
through  C'A,  B'A,  and  the  centre  of  the  sphere, 
we  have  the  arcs  B"A,  C'A,  and  B'A,  C'A  tangents 
to  them.  Hence,  B"AC"  is  the  same  as  B'AC,  and 
consequently  the  same  as  BAC. 

562,  ScH. — To  draio  an  arc  of  a  great  circle 
which  shall  be  perpendicular  to  another ;  or^  what  Fig.  324. 

M  the  same  thinrj,  to  construct  a  right  spherical  angle.  Let  it  be  required  to  erect 
an  arc  of  a  great  circle  perpendicular  to  CAB  at  A,  Fig.  323.  Lay  off  from  A,  on 
the  arc  CAB,  a  quadrant's  distance,  as  AP',  and  from  P'  as  a  pole,  with  a  quad- 
rant describe  an  arc  passing  through  A.    This  will  be  the  perpendicular  required. 

In  a  similar  manner  we  may  let  fall  a  perpendicular  from  any  point  in  the 
surface,  upon  any  arc  of  a  great  circle.  To  let  fall  a  perpendicular  from  P"  upon 
the  arc  CAB,  from  P"  as  a  pole,  with  a  quadrant  describe  an  arc  cutting  CAB, 
as  at  P'.  Then  from  P'  as  a  pole,  with  a  quadrant  describe  an  arc  passing 
through  P"  and  cutting  CAB,  and  it  will  be  perpendicular  to  CAB.  [The  stu- 
dent should  have  practice  in  making  these  constructions  on  the  sphere.] 


PROPOSITION  IX. 

5GS,   Problem* — To  pass  the  circumference  of  a  small  circle 
through  any  three  points  on  the  surface  of  a  sjyhere. 

Solution. — Let  A,  B,  and  C  be  the  three  points  in  the  surface  of  the  sphere 
through  which  we  propose  to  pass  the  circumference  of 
a  circle.  Pass  arcs  of  great  circles  through  the  points, 
forming  the  spherical  triangle  ABC.  Thus,  to  pass  an 
arc  of  a  iireat  circle  through  B  and  C,  from  B  as  a  pole, 
with  a  quadrant  strike  an  arc  as  near  as  may  be  to  the 
pole  of  the  required  circle  ;  and  from  C  as  a  pole,  with 
the  quadrant  strike  an  arc  intersecting  the  former,  as  at 
P  ;  then  is  P  the  pole  of  a  great  circle  passing  through 
B  and  C  (?).  Hence,  from  P  as  a  pole,  with  a  quadrant 
pass  an  arc  through  B  and  C,  and  it  will  be  the  arc  re- 
quired {551).  In  like  manner  pass  arcs  through  A  and 
C,  A  and  B.      Now,  bisect  two  of  these  arcs,  as  BC  and  AC,  by  arcs  of  great 


Fig.  325. 


218 


ELEMENTARY  SOLID  aEOMETRY. 


circles  perpendicular  to  each.  [The  student  will  readily  perceive  how  this  is 
done.]  The  intersection  of  these  perpendiculars,  o,  will  be  the  pole  of  the  small 
circle  required  (?).  Then  from  t>,  as  a  pole,  with  an  arc  oB  draw  the  circum- 
ference of  a  small  circle  :  it  will  pass  through  A,  B,  and  C  (?),  and  hence  is  the 
circumference  required. 


OF  TANGENT  PLANES. 

564z,  A  Tanffcnt  I^lane  to  a  curved  surface  at  a  given  point 
is  the  plane  of  two  lines  respectively  tangent  to  two  plane  sections 
through  the  point. 

III. — Let  P  be  a  point  in  the  cui*ved 
surfiice  at  which  we  wish  a  tangent 
plane.  Pass  any  two  planes  through 
the  surface  and  the  point  P,  and  let  OPQ 
and  MPN  represent  the  intersections  of 
these  planes  with  the  cui*ved  surface. 
Draw  UV  and  ST  in  the  planes  of  the 
sections,  and  tangent  to  OPQ  and  MPN, 
at  P.  Then  is  the  plane  of  UV  and  ST 
the  tangent  plane  at  P. 


Fig.  326. 


PROPOSITION  X. 

565.  TIieorein.—A  tangent  plane  to  a  sphere  is  perpendicular 
to  the  radius  at  the  point  of  tanrjency. 

Dem. — Let  P  be  any  point  in  the  surface 
of  a  sphere ;  pass  two  great  circles,  as  PaA, 
etc.,  and  PwAR,  through  P,  and  draw  ST 
tangent  to  the  arc  7nP,  and  UV  tangent  to  the 
arc  aP ;  then  is  the  plane  SVTU  a  tangent 
plane  at  P,  and  perpendicular  to  the  radius 
OP.  For,  a  tangent  (as  ST)  to  the  arc  mP  is 
perpendicular  to  the  radius  of  the  circle,  i.  e.y 
to  OP,  and  also  a  tangent  (as  VU)  to  the  arc  aP 
is  perpendicular  to  the  radius  of  this  circle, 
i.e.,  to  OP.  Hence,  OP  is  perpendicular  to 
two  lines  of  the  plane  SVTU.  and  conse- 
quently to  the  plane  of  these  lines  (?). 
q.  E.  D. 

566.  Cor.  l.~Every  point  in  a  tangeiit  plane  to  a  sphere,  except 
the  point  of  tangency^  is  without  the  sphere. 


OF  SPHERICAL  TRIANGLES.  219 

For,  OP,  the  perpendicular,  is  shorter  than  any  line  which  can  be  drawn 
from  0  to  any  other  point  in  the  plane  (?),  hence  any  other  point  in  the  plane 
than  P  lies  farther  from  the  centre  of  the  sphere  than  the  length  of  the  radius, 
and  is,  therefore,  without  the  sphere. 

567,  Cor.  2. — A  tangent  through  p  to  any  circle  of  the  sphere 
passing  through  this  point,  lies  in  the  tangent  pilane. 

Dem. — Thus  MN,  tangent  to  the  small  circle  PriRh  through  P,  lies  in  the 
tangent  plane.  For,  conceive  the  plane  of  the  small  circle  extended  till  it  in- 
tersects the  tangent  plane.  This  intersection  is  tangent  to  the  small  circle, 
since  it  touches  it  at  one  point,  but  cannot  cut  it;  otherwise  the  tangent  plane 
would  liave  another  point  than  P  common  with  the  surface  of  the  sphere.  But 
there  can  be  only  one  tangent  to  a  circle  at  a  given  point.  Hence  this  intersec- 
tion is  MN,  which  is  consequently  in  the  tangent  plane. 


OF  SPHERICAL  TRIANGLES. 

568,  A  Spherical  Trianffle  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  three  arcs  of  great  circles.  In  the  present  treatise 
these  arcs  will  be  considered  as  each  less  than  a  semicircumfer- 
ence. 

The  terms  scalene,  isosceles,  equilateral,  right  angled,  and  oblique 
angled,  are  applied  to  spherical  triangles  in  the  same  manner  as  to 
plane  triangles. 


PROPOSITION  XI^ 

569,  Theorem,— Tlie  sum  of  any  t200  sides  of  a  spherical  tri- 
angle is  greater  than  the  third  side,  and  their  difference  is  less  than 
the  third  side. 

Dem.— Let  ABC  be  any  spherical  triangle;  then  is 
BC  <  BA  +  AC,  and  BC  -  AC  <  BA ;  and  the  same  is 
true  of  the  sides  in  any  order.  For,  join  the  vertices  A, 
B,  and  C,  with  the  centre  of  the  sphere,  by  drawing  AO, 
BO,  and  CO.  There  is  tlius  formed  a  triedral  0-ABC, 
whose  facial  angles  are  measured  by  the  sides  of 
the  triangle  (208).  Now,  angle  BOC  <  BOA  +  AOC 
{4S4),  whence  BC  <  BA  +  AC  :  and  subtracting  AC 
from  both  members,  we  have  BC  —  AC  <  BA.  Fig.  3-28. 


220  ELEMENTARY  SOLID   GEOtLTRY. 


PROPOSITION  XIL 

570»  Tlieoretn, — The  sum  of  the  sides  of  a  spherical  triangle 
may  he  anytliing  between  0  and  a  "circumference. 

Dem. — The  sides  of  a  spherical  triangle  are  measures  of  the  facial  angles  of  a 
triedral  whose  vertex  is  at  the  ccntj-e  of  the  sphere.  Hence  their  sum  may  be 
anything  between  0  and  the  measure  of  4  right  angles,  as  these  are  the  limits 
of  the  sum  of  the  facial  angles  of  a  tiledral  {436). 

571*  ScH. — As  the  sides  of  a  spherical  triangle  are  arcs,  they  can  be  meas- 
ured in  dagrees.  Hence,  we  speak  of  the  side  of  a  spherical  triangle  as  30°, 
57",  1x5°  10',  etc.  In  accordance  with  this,  we  say  that  the  limit  of  the  sum  of 
;Le  sides  of  a  spherical  triangle  is  360°. 


PROPOSITION  xin. 

572*  Hieorem* — The  sum  of  the  atigles  of  a  spherical  triangle 
may  be  anything  between  tivo  and  six  right  angles. 

Dem. — The  sum  of  the  angles  of  a  spherical  triangle  is  the  same  as  the  sum 
of  the  measures  of  the  diedrals  of  a  triedral  having  its  vertex  at  the  centre  of 
the  sphere,  as  in  {569).  Now  the  limits  of  the  sum  of  the  measures  of  these 
diedrals  are  2  and  6  right  angles  {439).  Hence  the  sum  of  the  angles  of  any 
spherical  Uiangle  may  be  anything  between  2  and  6  right  angles.    Q.  e.  d. 

573,  ScH. — It  will  be  observed,  that  the  sum  of  the  angles  of  a  spherical 
triangle  is  not  constant,  as  is  the  sum  of  the  angles  of  a  plane  triangle.  Thus, 
the  sum  of  the  angles  of  a  spherical  triangle  may  be  200°,  290°,  350°,  500°,  any- 
thing between  180°  and  540°. 

574:.  Def. — Spherical  Excess  is  the  amount  by  which  the 
sum  of  the  angles  of  a  spherical  triangle  exceeds  the  sum  of  the 
angles  of  a  plane  triangle;  i.  e.,  it  is  the  sum  of  the  spherical  angles 
-180°,  or  7t. 

III. — It  is  not  difficult  to  observe  the  occasion  of  \hm  excess  in  the  case  of  the 
equilateral  spherical  triangle.  Thus.let  ABC  be  such  a  triangle.  Conceive  the  plane 

triangle  formed  by  the  chords  AB,  AC,  and  CB. 
The  sum  of  the  angles  of  this  plane  triangle  is 
180°.  Bat  each  angle  of  the  spherical  triangle 
is  larger  than  the  corresponding  angle  of  the 
plane  triangle.  Thus,  the  spherical  angle  BACis 
the  same  as  the  plane  angle  CAB',  included  be- 
tween the  tangents  C'A  and  B'A,  which  are  per- 
pendicular to  the  edge  of  the  diedral  C-AO-B,and 
Fia.  329.  include  its  measuring  angle.     Now,  CA  and  BA 


OF  SPHERICAL  TRIANGLES.  221 

being  different  line^  iro^i  C'A  and  B'A  are  oblique  to  the  edge  AO,  and  in- 
clude an  angle  less  than  its  measure,  and  consequently  less  than  CAB.  For 
a  like  reason  the  plane  angle  ACB  <  the  spherical  angle  ACB,  and  plane  angle 
ABC  <  spherical  angle  ABC.  Moreover,  it  is  easy  to  see  that  the  inequality 
between  -any  plane  angle  and  the  corresponding  spherical  angle  increases  as  the 
chords  BA  and  CA  deviate  more  from  the  tangents.  Whence  we  see  why  the 
sum  of  the  angles  of  the  spherical  triangle  is  not  a  fixed  quantity. 

575,  Cor. — A  spherical  triangle  may  have  one^  tioo,  or  even  three 
right  angles ;  and,  in  fact,  it  may  have  one,  t2uo,  or  three  obtuse 
angles  ;  since,  in  the  latter  case,  the  sum  'of  the  angles  luill  not  neces- 
sarily he  greater  than  540°. 

576.  Def.— ^  Trirectangular  Spherical  Triangle  is 

a  spherical  triangle  which  has  three  right  angles. 


PROPOSITION  XIV. 

577,  Theorein, — The  trirectangular  triangle  is  one-eighth  of 
the  surface  of  a  sphere. 

Dem. — Pass  three  planes  through  the  centre  of  a  sphere,  respectively  per- 
pendicular to  each  other.  They  will  divide  the 
surface  into  8  trirectangular  triangles,  any  one  of 
which  may  be  applied  to  any  other.  Thus,  let 
ABA'B',  ACA'C,  and  CBC'B'  be  the  great  circles 
formed  by  the  three  planes,  mutually  perpendicu- 
lar to  each  other.  The  planes  being  perpendicular 
to  each  other  the  diedrals,  as  A-CO-B,  C-BO-A, 
C-AO-B,  etc.,  are  right,  and  hence  the  angles  of 
the  8  triangles  formed  are  all  right.  Also,  as  AOB 
is  a  right  angle,  AB  is  a  quadrant;  as  BOC  is  a 
right  angle,  CB  is  a  quadrant,  etc.  Hence,  each 
side  of  every  triangle  is  a  quadrant.  Now  any  one  triangle  may  be  applied  to 
any  other.  [Let  the  student  make  the  application.]  Hence  the  trirectangular 
triangle  is  one-eighth  of  the  surface  of  a  sphere.    Q.  e.  d. 

,  57 S»  Cor. — Tlie  trirectangular  triangle  is   equilateral  and  its 
sides  are  quadrants. 


PROPOSITION   XV. 
579,  TJieorem* — In  an  isosceles  spherical  tria^igle  the  angles 
opposite  the  equal  sides  are  equal  j  and,  conversely.  If  two  angles  of 
a  spherical  triangle  are  equal,  the  triangle  is  isosceles. 


222 


ELEMENTARY  SOLID  GEOMETRY. 


Dem. — Let  ABC  be  an  isosceles  spherical  triangle  in  which  AB  =  AC  ;  then 
angle  ABC  =  ACB.  For,  draw  the  radii  AO,  CO,  and 
BO,  forming  the  edges  of  the  triedral  O-ABC.  Now, 
since  AB  =  AC,  the  facial  angles  AOC  and  AOB  are 
equal,  and  the  triedral  is  isosceles.  Hence  the  dic- 
drals  A-OB-C  and  A-OC-B  are  equal  {442),  and  con- 
sequently the  spherical  angles  ABC  and  ACB  are 
equal  {558).  Again,  if  angle  ABC  =  angle  ACB,  side 
AC  =  side  AB.  For  in  the  triedral  O-ABC,  the  die- 
drals  A-OB-C  and  A  -OC-B  are  equal,  whence  the  facial 
angles  AOB  and  AOC  are  equal  {443),  and  conse- 
quently the  sides  AB  and  AC  which  measure  these  angles. 


Fig.  331. 


580,  Cor. — An  equilateral  spherical  triangle  is  also  equiangular  ; 
and,  conversely,  If  the  angles  of  a  splierical  triangle  are  equal  the 
triangle  is  equilateral. 


^ 


PROPOSITION  XVI. 


SSI.  Hieorem. — On  the  same  or  07i  equal  spheres  tivo  isosceles 
triangles  having  two  sides  and  the  included  angle  of  the  one  equal  to 
two  sides  and  the  included  angle  of  the  other,  each  to  each,  can  he 
superimposed,  and  are  consequently  equal, 

Dem.— In  the  triangles  ABC  and  AB'C,  let  AB  =  AC,  AB'  =  AC ;  and  let 
AB  =  AB',  BC  =  B'C,  and  angle  ABC  =  ABC ;  then 
can  the  triangle  AB'C  be  superimposed  upon  ABC. 
For,  since  the  triangles  are  isosceles,  we  have  angle  ABC 
—  ACB,  AB'C  =  ACB',  and,  as  by  hypothesis  ABC  = 
AB'C,  these  four  angles  are  equal  each  to  each.  For  a 
like  reason  AB  =  AC  =  AB'  =  AC.  Now,  applying 
AC  to  its  equal  AB,  the  extremity  A  at  A  and  C  at  B, 
with  the  angle  B'  on  the  same  side  of  AB  as  C,  the  con- 
vexities of  the  arcs  AC  and  AB  being  the  same,  and  in 
the  same  direction,  the  arcs  will  coincide.  Then,  as 
angle  ACB'  =  ABC,  CB'  will  take  the  direction  BC,  and  since  these  arcs  are 
equal  by  hypothesis,  B'  will  fall  at  C.  Hence  B'A  will  fall  in  CA.  as  only  one 
arc  of  a  great  circle  can  pass  between  C  and  A,  and  the  triangle  AB'C  is  super- 
imposed upon  ABC;  wherefore  they  are  equal.  [Let  the  student  give  the 
application  when  other  parts  are  assumed  equal.] 


582,  Symmetmcal  Spherical  Triangles  are  such  as 
have  the  parts  (sides  and  angles)  of  the  one  respectively  equal  to  the 
parts  of  the  other,  but  arranged  in  a  dijfferent  order,  so  that  the  tri- 
angles are  not  capable  of  superposition. 


OF  SPHERICAL  TRIANGLES. 


223 


Fia.  :WJ. 


III.— In  Fig.  333,  ABC  and  A'B'C  represent  sj-mmetiical  spherical  tri- 
angles. In  these  triangles  A  =  A',  B  =  B',  C  =  C, 
AC  =  A'C^  AB  =  A'B',  and  BC  =  B'C ;  neverthe- 
less we  cannot  conceive  one  triangle  superimposed 
upon  the  other.  Thus,  were  we  to  make  the  at- 
tempt by  placing  A'B'  in  its  equal  AB,  A'  at  A,  and 
B'  at  B,  the  angle  C  would  fall  on  the  opposite  side 
of  AB  from  C.  Now,  we  cannot  revolve  A'C'B'  on 
AB  (or  its  chord),  and  thus  make  the  two  coincide, 
for  this  would  bring  their  convexities  together. 
Nor  can  we  make  them  coincide  by  reversing  A'B'C, 
and  placing  B'  at  A,  and  A'  at  B.  For,  although 
these  two  arcs  will  thus  coincide,  as  the  angle  B'  is 
not  equal  to  A,  B'C  will  not  fall  in  AC ;  and,  again, 
if  it  did,  C  would  not  fall  at  C,  since  B'C  and  AC  are 
not  equal. 

But,  considering  the  triangles  ABC  and  A'B'C  in 
Fig.  334,  in  which  A  =r  A',  B  =  B',  C  =  C,  AC  = 
A'C,  AB  =:  A'B',  and  BC  =  B'C,  we  can  readily 
conceive  the  latter  as  superimposed  upon  the  former. 
[The  student  should  make  the  application.]  Now, 
the  two  triangles  are  equal  in  each  case,  as  will 
subsequently  appear  of  the  former.  Such  triangles  as 
tliose  in  Fig.  333  are  called  symmetHcally  equal,  while 
the  latter  are  said  to  be  equal  by  superposition. 

Fig.  335  represents  the  same  triangles  as  Fig.  334, 
and  exhibits  a  complete  projection*  of  the  semicir- 
cumferences  of  which  the  sides  of  the  triangles  are 
arcs.  The  student  should  become  perfectly  familiar 
with  it,  and  be  able  to  draw  it  readily.  Thus,  a^Bh 
is  the  projection  of  the  semicircumference  of  which  Fig.  a35. 

AB  is  an  arc,  aACc  of  the  semicircumference  of  which  AC  is  an  arc,  etc.,  etc. 


Fig.  334. 


PROPOSITION  XTII. 

S83,  Theorem, — Symmetrical  spherical  triancjUs  are  equiva- 
lent. 


'*  To  understand  what  is  meant  by  the  projection  of  these  lines,  conceive  a  hemisphere 
irith  its  base  on  the  paper,  and  represented  by  the  circle  abc,  and  all  the  arcs  raised  up  from  the 
paper  as  they  would  be  on  the  sui-fe,ce  of  such  a  hemisphere.  Thus,  considering  the  arc  aABft, 
the  ends  a  and  b  would  be  in  the  paper  jutt  where  they  are,  but  the  rest  of  the  arc  would  be 
off  the  paper,  as  though  you  could  take  hold  of  B  and  raise  it  from  the  paper  while  a  and  b 
remain  fixed.  The  lines  in  the  figure  are  representations  of  lines  on  the  surface  of  such  a 
hemisphere,  as  they  would  appear  to  an  ej'e  situated  in  the  axis  of  the  circle  abc,  and  at  an 
infinite  distance  from  it;  that  is,  just  as  if  each  point  in  the  lines  dropped  7>er/)««(/ic«^ar/y 
down  upon  the  paper.  Arcs  of  great  circles  perpendicular  to  the  base  are  projected  in  straight 
lines  passing  through  the  centre,  and  oblique  arcs  are  projected  in  ellipses.  See  Spherical 
Trigonometry  {97-/09), 


224 


ELEMENTARY  SOLID  GEOMETRY. 


Dem. — Let  ABC  and  A'B'C  be  two  symmetrical  spherical  triangles,  with  AB 
=  A'B',  AC  =  A'C\  BC  =  B'C,  A  =  A',  B  =  B',  and  C  =  C ;  then  are  they 
equivalent 

For,  pass  circumferences  of  small  circles  through  the 
vertices  A,  B,  C  and  A',  B',  C,  as  abc  and  a'b'c\  of  which 
o  and  o'  are  the  poles.  [The  student  should  execute  this 
on  the  spherical  blackboard.]  Now,  by  reascm  of  the 
mutual  equality  of  the  sides,  the  chord  AC  =  chord  A'C, 
chord  AB  =  cJiord  A'B\  and  chord  BC  =  cJiord  B'C,  and 
as  the  small  circles  are  circumscribed  about  the  equal 
plajie  triangles  ABC  and  A'B'C,  these  circles  are  equal. 
Hence,  oA  =  o'A'  =  oB  =  o'B'  =  oC  =  o'C.  The  tri- 
angle AoB  is  therefore  equal  to  At^'B',  BoC  =  B'o'C,  and 
A<?C  =  AVC.  [The  student  should  make  the  application  of  these  equal  tri- 
angles.] Hence,  ABC  is  equivalent  to  A'B'C,  as  the  two  are  composed  of  equal 
parts. 

Kthe  poles  of  the  small  circles  fell  without  the  given  triangles,  ABC  would 
be  equivalent  to  the  sum  of  two  of  the  partial  triangles  minus  the  third. 


Fig.  336. 


Fig.  337. 


PROPOSITION   XTni. 

584,  Theorem, — On  the  same  or  equal  sj^lieres,  tivo  spJier-ical 

triangles  having  two  sides  and  the  included  angle 
of  the  one  equal  to  tivo  sides  and  the  included 
angle  of  the  other,  each  to  each,  are  equal,  or  sym^ 
metrical  and  equivalent. 

Dem.— Let  ABC  and  A'B'C,  Fig.  n37,  be  two  spherical 
triangles  having  AB  =  A'B',  AC  =  A'C,  nnd  A  =  A'.  In 
this  case,as  the  parts  are  similarly  arranged,  by  placing  AC 
in  its  equal  A'C,  AB  will  fall  in  its  equ.-il  A'B'  (as  A  =  A'"), 
and  the  two  triangles  will  coincide.  Hence,  tliey  are  equal. 
Again,  let  the  two  triangles  be  ABC  and  A'B'C,  Fig.  338, 
in  which  AB  =  A'B',  AC  =  A'C,  and  A  =  A',  the  parts 
not  being  similarly  arranged,  so  tluit  the  triangles  are 
incapable  of  superposition.  Thus,  if  A3  is  placed  in  its 
equal  A'B',  A  at  A',  and  B  at  B',  C  and  C  will  fall  on 
opposite  sides  of  AB.  We  maj%  however,  construct  ABC, 
Fig.  337,  symmetrical  with  A'B'C  in  this  figure,  and  ap- 
ply ABC  oiFig.  338  to  it,  and  find  that  they  coincide. 
Now,  ABC,  Fig.  337,  and  A'B'C,  Fig.  338,  are  equivalent 
Fig.  338.  {58:^) ;  hence  ABC,  Fig.  338,  is  equivalent  to  A'B'C, 

Fig.  338. 

585.  Sen.— This  proposition  is  virtually  the  same  as  {440)  concerning  trie- 
drals.   Thus,  in  Fig.  337,  drawing  the  radii  AO,  BO,  CO,  A'O,  B'O,  and  CO,  two 


OF  SPHERICAL  TRIANGLES.  225 

triedrals  are  formed,  having  the  facial  angle  AOB  =  A'OB  ,  AOC  =  A'OC, 
the  included  diedrals  equal,  and  the  parts  similarly  disposed,  whence  the  trie- 
drals are  equal.  In  like  manner  the  triedral  0-ABC,  Fe^r.  338,  is  symmetrical 
and  equivalent  to  O-A'B'C,  Fig.  338.  Hence,  in  either  case,  all  the  parts  of 
one  spherical  triangle  are  equal  to  all  the  parts  of  the  other,  each  to  each. 


PROPOsiTiox  xrx. 

586.  Theorem, — On  the  same,  or  07i  equal  spheres,  tiuo  spher^'- 
cdl  triangles  having  tioo  angles  and  the  included  side  of  the  one  equal 
to  tivo  angles  and  the  included  side  of  the  other,  each  to  each,  are 
equal,  or  sy7nmetrical  and  equivalent. 

Dem. — Using  the  same  triangles  as  in  the  preceding  proposition,  the  student 
should  be  able  to  make  the  application  directly,  when  the  parts  are  similarly 
disposed ;  and  when  not  similarly  disposed,  he  should  be  able  to  show  that 
ABC,  of  Fig.  338,  can  be  applied  to  ABC,  Fig.  337,  symmetrical  with  A'B'C, 
Fig.  338. 

587.  ScH. — This  proposition  is  also  yirtually  the  same  as  {,447)  concerning 
triedrals.     Let  the  student  point  out  the  identity. 


PROPOSITION   XX. 

S88,  TJieovem. — On  the  same,  or  on  equal  spheres,  if  tivo 
spherical  triangles  have  two  sides  of  the  one  equal  to  two  sides  of 
the  other,  each  to  each,  and  the  included  angles  unequal,  the  third 
sides  are  unequal,  and  the  greater  third  side  belongs  to  the  triangle 
having  the  greater  included  angle.  Conversely,  If  the  two  sides  are 
equal,  each  to  each,  and  the  third  sides  unequal,  the  angles  included 
hy  the  equal  sides  are  unequal,  and  the  greater  belongs  to  the  triangle 
having  the  greater  third  side. 

'Dem.— In  the  triangles  ABC  and  A'B'C,  let  AB  =  A'B', 
AC  =  A'C,  and  A  >  A' ;  then  is  BC  >  B'C.  For,  join  the 
vertices  with  the  centre,  forming  the  two  triedrals  0-ABC 
and  O-A'B'C.  In  these  triedrals  AOB  =  A'OB',  AOC 
=  A'OC,  being  measured  by  equal  arcs ;  and  C-AO-B 
>  C'-A'O-B',  having  the  same  measures  as  A  and  A'  {558). 
Hence  COB  >  COB'  {449).  Therefore  CB,  the  measure 
of  COB,  is  greater  than  CB',  the  measure  of  COB'. 

In  like  manner,  the  same  sides  of  the  triangles,  and  con-  Fig.  339. 

sequently  the  same  facial  angles  of  the  triedrals,  being  granted  equal,  and 

15 


226  ELEMENTARY  SOLID  GEOMETEY. 

BC  >  B C,  A  >  A'.    For,  BC  being  greater  tlian  B'C,  COB  >  COB' ;  whence 
B-AO-C  >  B'-A'C-C  {450),  or  A  is  greater  than  A'. 


PROPOSITION  XXI. 

SS9,  Hieoreni, — On  the  same,  or  on  equal  spheres,  iwo  spheri- 
cal triangles  having  the  sides  of  the  one  respectively  equal  to  the  sides 
of  the  other,  or  the  a?igles  of  the  one  respectively  equal  to  the  angUs 
of  the  other,  are  equal,  or  symmetrical  and  equivalent. 

Dem. — The  sides  of  the  triangles  being  equal,  the  facial  angles  of  the  triedrale 
at  the  centre  are  equal,  whence  the  triedrals  are  equal  or  symmetrical  {451). 
Consequently  the  angles  of  the  triangles  are  equal,  and  the  triangles  are  equal, 
or  symmetrical  and  equivalent. 

Again,  the  tiiangles  being  mutually  equiangular,  the  triedrals  have  their 
diedrals  mutually  equal ;  whence  the  triedrals  are  equal  or  symmetrical  {452). 
Therefore,  the  sides  of  the  triangles  are  mutually  equal,  and  the  triangles  ai'C 
equal,  or  symmetrical  and  equivalent.    (See  Figs.  333,  334) 


PROPOSITION  xxn. 

S90,  Tlieorem, —  On  spheres  of  different  radii,  mutually  equi- 
angular  triangles  are  similar  (not  equal). 

Dem. — Let  0  be  the  common  centre  of  two  un- 
equal spheres  ;  and  let  ABC  be  a  spherical  triangle 
on  the  surface  of  the  outer.  Draw  the  radii  AO,  BO, 
and  CO,  constructing  the  triedral  0-ABC.  Now, 
the  intersections  of  these  faces  with  the  surface  of 
the  inner  sphere  will  constitute  a  ti-iangle  which  is 
mutually  equiangular  with  ABC.  Thus,  A  r=  «, 
B  =  6,  and  C  =  c,  since  in  each  case  the  correspon- 
ding diedrals  are  the  same.  From  the  similar  sec- 
tors aOh,  AOB,  we  have  aZ>  :  AB  : :  aO  :  AO;  and, 
Fis.  340.  in  like  manner,  oc  :  AC  : :  aO  :  AO.     "Whence,  ab  : 

AB  :  :  flrc  :  AC.  So,  also,  ab  :  AB  :  :  iO  :  BO,  and 
be:  BC::  50  :  BO ;  whence,  ab  :  AS  : -.  be  :  BC.  Thus  we  see  that  ABC  and 
abc,  having  their  angles  equal  each  to  each,  have  also  their  sides  proportional: 
therefore  they  are  similar. 


POLAR  OR  SUPPLE3IENTAL  TRIANGLES. 

S91»  One  triangle  is  polar  to  another  when  the  vertices  of  one 
are  the  poles  of  the  sides  of  the  other.     Such  triangles  are  also 


POLAR  OR  SUPPLEMENTAL  TRIANGLES. 


227 


called  supplemental,  since  the  angles  of  one  are  the  supplements  of 
the  sides  opposite  in  the  pther,  as  will  appear  hereafter. 


PROPOSITION  XXIII. 


592,  I^rohlem^ — Having  a  spherical 
triangle  given,  to  draw  its  i^olar. 

Solution.— Let  ABC  be  the  given  triangle  *  From 
A  as  a  pole,  with  a  quadrant  strike  an  arc,  as  C'B', 
From  B  as  a  pole,  with  a  quadrant  strike  the  arc 
C'A' ;  and  from  C,  the  arc  A'B'.  Then  is  A'B'C 
polar  to  ABC. 

59H,  Cor. — If  one  triangle  is  polar  to 
anotlier,  conversely,  the  latter  is  polar  to  the 
former  ;  i.  e.,  the  relation  is  reciprocal. 


Fig.  341. 


Thus,  A'B'C  being  polar  to  ABC  ;  reciprocally,  ABC  is  polar  to  A'B'C;  that 
is,  A'  is  the  pole  of  CB,  B'  of  AC,  and  C  of  AB.  For  every  point  in  A'B'  is 
at  a  quadrant's  distance  from  C,  and  every  point  in  A'C  is  at  a  quadrant's  dis- 
tance from  B.  Hence,  A'  is  at  a  quadrant's  distance  from  the  two  points  C  and 
B  of  CB,  and  is  therefore  its  pole.  [In  like  manner  the  student  should  show 
that  B'  is  the  pole  of  AC,  and  C  of  AB.] 


«5i>4.  Sen. — By  producing  each  of  the  arcs 
struck  from  the  vertices  of  the  given  triangles 
suflaciently,/<?Mr  new  triangles  will  be  formed,  viz., 
A'B'C,  QC'B',  PC'A',  and  RA'B'.  Only  the  lirst 
of  these  is  called  polar  to  the  given  triangle. 
It  is  easy  to  observe  the  relation  of  any  of  the 
parts  of  any  one  of  the  other  three  triangles  to 
the  parts  of  the  polar.  Thus,  QC  =  180°  —  b', 
QtB'  =  180°  -  c',  QC'B'  =  180°  -  B'C'A',  QB'C 
=  180°  -  C'B'A',  and  Q  =  A'  =  180°  -  a,  as  will 
appear  hereafter. 


P< 


Fig.  342. 


*  This  should  be  executed  on  a  sphere.  Few  students  get  clear  ideas  of  polar  triangles 
without  it.  Care  should  be  taken  to  construct  a  variety  of  trianp^Ies  as  the  given  triangle, 
since  the  polar  triangle  does  not  always  lie  in  the  position  indicated  in  the  figure  here  given. 
Let  the  given  triangle  have  one  side  considerably  greater  than  90°,  another  somewhat  less, 
and  the  third  quite  small.  Also,  let  each  of  the  sides  of  the  given  triangle  be  greater 
than  90°. 


228 


ELEMENTAEY  SOLID  GEOMETRY. 


Fig.  ai3. 


PROPOSITION  XXIY. 

o96»  Tlieorem, — Amj  axgle  of  a  spherical  triangle  is  the 
svppJement  of  the  side  opposite  in  its  polar  triangle  ;  and  any  side 
is  the  supplement  of  the  angm:  opposite  in  the  polar  triangle. 

Dem.— Let  ABC  and  A'B'C  be  two  spherical  tri- 
angles polar  to  each  other ;  and  let  the  sides  of 
each  be  designated  as  a,  6,  c,  a\  b',  c\  a  being 
opposite  A,  a'  opposite  A',  h  opposite  B,  etc.  Then 
A  =  180°  -a\B  =  180°  -h\0  =  180°  -  c',  a  = 
180°  -  A',  6  =  180°  -  B',  and  c  =  180°  -  C 

For,  join  the  vertices  of  the  triangles  with  the 
centre  of  the  sphere,  thus  forming  the  triedrals 
0-ABC,  and  O-A'B'C.  These  triedrals  are  sup- 
plemental ;  for,  A  being  the  pole  of  C'B',  AO  is  the 
axis  of  the  great  circle  of  which  C'B'  is  an  arc  (?), 
hence  is  perpendicular  to  the  plane  COB',  and 
consequently  to  OB'  and  OC  (?).  In  like  manner, 
BO  is  perpendicular  to  the  plane  A'OC,  and  hence  to  OA'  and  OC.  So,  also, 
CO  is  perpendicular  to  OA'  and  OB'.  Now,  these  triedrals  being  supplement- 
ary, thediedral  B-AO-C  is  the  supplement  of  the  facial  angle  COB'  (438);  or, 
since  the  diedral  B-AO-C  is  the  same  as  the  spherical  angle  A,  and  the  facial 
angle  COB'  is  measured  by  a\  A  is  the  supplement  of  «',  i.  e.,  A  =  180°  —  a'. 
For  like  reasons,  B  =  180°  —  b\  and  C  =  180°  —  c'.  [Let  the  student  give  them 
in  full]  Again,  the  diedral  B'-A'O-C  is  the  supplement  of  the  facial  angle 
COB  {438);  whence  A'  =  180°  -  a.  In  like  manner  B'  =  180°  -  b,  and  C  = 
180°  -  c. 

Second  Demonstration. — Let  ABC  and  A'B'C  be  two 
polar  triangles.  Let  CB,  CA,  and  AB  be  represented  by  a, 
b,  and  c  respectively,  and  CB',  CA',  and  A'B'  by  a',  b',  and  c'. 
To  show  that  A  =  180°  —  «',  produce  b  and  c,  if  necessaiy,  till 
they  meet  the  side  a',  of  the  triangle  polar  to  ABC,  in  e  and 
d.  Now  A  is  measured  by  ed  {560).  But,  since  Z'd  —  90°, 
and  B'e  =  90°,  Z'd  +  B'^,  or  CB'  +  €d  =  180° ;  whence  trans- 
posing, and  putting  a'  for  CB',  we  have  ed  =  k  =  180°  —  a'. 
In  like  manner  Z'g  +  A'/=  CA'  +/^  =  180° ;  whence  //;  =  B  =  180°  -  CA', 
or  180°  -  6'.  So,  also,  C  =  180°  -  c'.  To  show  that  A'  =  180°  -  a,  consider 
that  A'  being  the  pole  of  CB,/i  is  the  measure  of  A'.  Now  B/=  90°  (?),  and 
C^  =  90° ;  whence  B/  +  Zi  -  180°.  But  B/  +  Ci-fi  +  «,  wherefore  fi  +  a  = 
180°,  and  transposing,  and  putting  A'  for /i,  we  have  A'  =  180°  -  a.  In  like  man- 
ner w^  may  show  that  B'  -  180°  -  6,  and  C  =  180°  -  c.  [The  student  should 
give  the  details.] 


Fia.  344. 


QUADRATURE  OF  SURFACE  OF  SPHERE.  229 


QUADRATURE  OF  THE  SURFACE  OF  THE  SPHERE. 

S96.  The  Quadrature  *  of  a  surface  is  the  same  as  finding  its 
area.  The  term  is  applied  under  the  conception  that  the  process 
consists  in  finding  a  square  which  is  equivalent  to  the  given  surface. 


PROPOSITION  XXY. 

5911,  Lem^na, — The  surface  generated  by  the  revolution  of  a 
regular  semi-2)olygon  of  an  even  number  of  sides,  about  the  diameter 
of  the  circumscribed  circle  as  an  axis,  is  equivalent  to  the  circum- 
ference of  the  inscribed  circle  multiplied  by  the  axis. 

Dem.— Let  ABCDE  be  one  half  of  a  regular  octagon,  AE  ^ 

being  the  diameter  of  the  circumscribing  circle.     If  the  semi-  <t^ 

perimeter  ABCDE  be  revolved  about  AE  as  an  axis,  the  surface  ^^^^--r-- 

generated  will  be  ^nr  x  AE,  r  being  the  radius  of  the  inscribed         3  / 1 \ 

circle,  as  «0,  or  hO.  y'  p-,,.^ 

This  surface  is  composed  of  the  convex  surfaces  of  cones       ^Vd ~ 

and  frustums  of  cones.    Thus  AB  generates  the  surface  of  a  \ 

cone,  BC  the  frustum  of  a  cone,  etc.    Let  a  and  b  be  the  mid-  d^ — 

die  points  of  AB  and  BC,  and  draw  am,  Be,  bn,  and  CO  per-  ^^^ 

pendicular  to  the  axis,  and  B^  parallel  to  it.    Also  draw  the 

radii  of  the  inscribed  circle,  aO  and  bO.     Indicate  the  sur-  Fig-  345. 

faces  generated  by  the  sides,  as    Surf.  AB,  Surf.  BC,  etc. 

The  areas  of  these  surfaces  are : 

Surf  AB=z27t  X  mi  x  AB  {5 J 6),  (1) 

Surf  BC  =  27t  X  bii  x  BC  {518),  etc.  (3) 

Now,  from  the  similar  triangles  0am  and  BAc, 
"We  have        aO  :  AB  : :  am  :  Ac,  or  2;r  x  «0  :  AB  : :  2;r  x  am  :  Ac  ; 
Whence         2;r  x  rtm  x  AB  =  27tr  x  Ac,  putting  r  for  aO. 

Also,  from  the  similar  triangles  Obn  and  CBfZ, 
We  have        bO  :  BC  : :  bn -.  Bd  {=  cO),  or  Stt  x  50  :  BC  : :  2;r  x  &/i  :  cO  ; 
Whence         27C  x  bn  x  BC  =  2itr  x  cO,  putting  r  for  bO. 

Substituting  these  values  in  (1)  and  (2),  we  obtain 

Surf  AB  =  27tr  x  Ac, 

Surf.  BC  =:  27tr  x  cO, 

And,  in  like  manner.  Surf.  CD  =  27rr  x  Op, 

And,  Surf  DE  =  27tr  x  pE. 


Adding,  Surf  ABCDE  =z  27tr  (Ac  +  cO  +  Op  +  pE)  =  2:rr  x  AE. 

Finally,  since  the  same  course  of  reasoning  is  applicable  to  the  semi-polygons 
of  16,  32,  64,  etc.,  sides,  the  truth  of  the  proposition  is  established. 

♦  Latin  quadratus,  squared. 


230  ELEMENTARY  SOLID  GEOMETRY. 

598,  ScH. — This  proposition  is  only  a  particular  case  of  surfaces  generated 
by  any  broken  line  revolving  about  an  axis ;  and  the  general  proposition  can  be 
established  in  a  manner  altogether  similar  to  the  method  given  above.  But  this 
case  is  all  that  we  need  for  our  present  purpose. 


PROPOSITION  XXYI. 

599,  Tlieorem, —  The  surface  of  a  sphere  is  equivalent  to  four 
great  circles  ;  that  is,  to  47rR''',  R  being  the  radius  of  the  sphere. 

Dem. — Let  the  semicircumference  ABCDE  revolve  upon 
the  diameter  AE,  and  thus  generate  the  surface  of  a  sphere. 
Conceive  the  half  of  a  regular  octagon  inscribed  in  the 
•emicircle.  Call  the  radius  of  the  inscribed  circle,  as  aO,  r, 
and  let  AO  =  R  The  surface  generated  by  the  broken  line 
ABCDE  is,  by  the  last  proposition,  27tr  x  2  R  =  4;rrR  Now, 
conceive  the  arcs  AS,  BC,  etc.,  bisected,  and  the  chords  drawn, 
and  let  ?*'  bQ  the  radius  of  the  circle  inscribed  in  the  regular 
polygon  thus  formed.  The  surface  generated  by  this  semi- 
polygon  will  be  A.ifr''R.  By  repeating  the  bisections,  the 
^^'      '  broken  line  approximates  to  the  semicircumference,  the  radius 

of  the  inscribed  circle  to  R,  and  the  surface  generated  to  the  surface  of  the 
sphere,  the  three  quantities  reaching  their  limits  at  the  same  time.  Hence  at 
the  limit  we  have 

Surf,  of  sphere  =  27rR  x  2R  =  47rR'.    q.  e.  d. 

OOO.  CoE.  1. — The  area  of  the  surface  of  a  sphere  ii  equivalent  to 
the  circumference  of  a  great  circle  multiplied  hy  the  diameter ,  that  is, 
2;rR  X  2R,  as  above, 

601.  Cor.  2. — The  surfaces  of  spheres  are  to  each  other  as  the 
squares  of  their  radii. 

Thus,  if  R  and  R'  are  the  radii  of  two  spheres,  the  sm*faces  are  4;rR'''  and 
4;rR'=.    Now,  4;rR='  :  47rR'^  : :  R^  :  R  ^ 


602,  Def. — A  Zone  is  the  portion  of  the  surface  of  a  sphere 
included  bet"\veen  the  circumferences  of  two  parallel  circles  of  a 
sphere.  The  altitude  of  a  zone  is  the  distance  between  the  parallel 
circles  forming  its  bases. 

III. — The  surface  generated  by  arc  CB,  or  €iny  arc  of  the  circle  ABCDE, 
Fig.  346,  in  its  revolution,  conforms  to  the  definition,  and  is  a  zone.  Such  a 
portion  of  the  surface  as  is  generated  by  A B  is  called  a  zone  with  one  base,  the 
circle  whose  circumference  would  form  the  upper  base  having  become  tangent 
to  the  sphere. 


OF  LUNES. 


231 


/PROPOSITION  xx\n. 

603,  Theorem, —  The  area  of  a  zone  is  to  the  area  of  the  surface 
of  the  sphere  as  the  altitude  of  the  zone  is  to  the  diameter  of  the 
sphere;  which  gives  for  the  area  of  a  zone  27tSiR,  a  being  the  altitude 
of  the  zone,  and  R  the  radius  of  the  sphere. 

Dem. — It  is  evident  that  in  passing  to  the  limit  the  surface  generated  by  such 
a  portion  of  the  broken  line  as  would  lie  between  C  and  B,  Fig.  346,  would 
be  measured  by  the  circumference  of  the  inscribed  circle  multiplied  by  cO. 
Hence,  at  the  limit,  the  zone  gc^nerated  by  arc  BC  is  measured  by  2;rR  x  cO,  that 
is,  it  is  such  a  part  of  the  surface  of  the  sphere  as  cO  is  of  AE,  or  3R.     liCtting  a 

represent  the  altitude  cO,  the  fraction  ^  represents  the  part  of  the  surface  of 


X  -—,  which  equals 


the  sphere  constituting  the  area  of  the  zone.    Hence,  4:7tK^ 
27tdR,  is  the  area  of  the  zcjLe. 

604,  CoK. — On  the  same  or  on  equal  spheres,  zones  are  to  each 
other  as  their  altitudes.  / 


} 


OF  LUNES. 


605,  A  Lune  is  a  portion  of  the  surface  of  a  sphere  included 
by  two  semicirc^'mferences  of  great  circles. 

The  smface  AmBw.  is  a  lune. 

606,  The  Angle  of  the  Lune  is  the  angle  in- 
cluded by  the  arcs  which  form  its  sides ;  or,  what 
is  the  same  thing,  the  measure  of  the  diedral  in- 
cluded between  the  great  circles. 

Thus,  the  spherical  angle  mkn,  or  the  measure  of  the 
diedi'al  m-AB-7i,  is  the  angle  of  the  lune  f<mBn. 


PROPOSITION  xxyni. 

60t,  Theorem. — The  area  of  a  lune  is  to  the  area  of  the  surface 
of  the  sphere  on  lohich  it  is  situated,  as  the  angle  of  the  lune  is  to 
four  right  angles. 

Dem. — Let  ACEB  be  a  mne  whose  angle  is  the  spherical  angle  CAB,  or  what 


232 


ELEMENTARY  SOLID  GEOi'ZETRY. 


is  tlie  same  thing,  the  plane  angle  BOC  measured 
by  the  arc  CB,  of  which'  A  is  the  pole ;  then  is 

lune  ACEB  :  surface  of  spCiere  : :  CAB  :  4  right  angles. 

For,  suppose  the  arc  CB  commensurable  with  the 
circumference  BCwD;i,  and  suppose  that  they  are 
to  each  other  as  5  :  24.  Dividing  BC  into  5  equal 
arcs,  and  the  entire  c  ii'cumference  BC7wD;i.  into  24 
arcs  of  the  same  length,  and  passing  arcs  of  great 
circles  through  A  and  these  points  of  division,  the 
lune  will  be  divided  into  5  equal  lunes,  and  the 
entire  surface  into  24^  equal  lunes  of  the  same  size. 
That  these  lunes  are  equal  to  each  other  is  evident  fi-om  the  fact  that  they  are 
composed  of  equal  isosceles  tiiangles.    Hence, 

lune  ACEB  :  surface  of  sphere  :  •-  5  :  24. 
Now,  angle   BOC  :  4  right  angles  : :  BC  (=  5-^  :  BC;7iD;i  (=  24). 

Therefore,      lune  ACEB  :  surface  of  sphere  ::  BOC  (^z*  CAB)  :  4:  right  angles^ 
since  the  circumference  measures  4  right  angles.  ' 

If  BC  has  no  finite  common  measure  with  the  circumference,  we  may  divide 
it  into  an}-  number  of  equal  arcs,  bisect  these  arcs,  ther.  bisect  the  last  formed, 
and  continue  the  process  of  bisection  (in  conception)  ta  any  required  extent ; 
and  as,  when  any  one  of  the  arcs  thus  obtained  is  applied  to  the  circumference, 
if  it  is  not  an  exact  measure,  the  remainder  is  less  than  tt  ^  arc,  we  can  continue 
the  subdi\'ision  of  BC  (in  conception)  until  this  remaii-der  is  less  than  any 
assignable  quantity.  Hence,  we  may  always  consider  *he  arc  BC  as  com- 
mensurable with  the  circumference  by  making  the  measure  infinitesimal. 

608,  Cor. — TJie  sum  of  several  lunes  on  the  same  sphere  is  equal 
to  a  June  whose  angle  is  the  sum  of  the  angles  of  the  lunes;  and  the 
difference  of  two  lunes  is  a  lune  whose  angle  is  the  diff^erence  of  their 
angles. 

609.  ScH.  1. — The  case  in  which  the  arc  measuring  the  angle  of  the  lune 
Is  incommensurable  with  the  circumference,  may  be  treated  as  in  {206),  by  the 
method  of  reasoning  called  the  Eeductio  ad  dbsurdum,  i.  e.,  by  showing  a  thing 
io  be  true,  since  it  would  be  absurd  to  suppose  it  untrue. 

Thus,  there  is  some  arc  to  which  the  circum- 
ference bears  the  same  ratio  as  the  surface  of  the 
sphere  does  to  the  surface  of  the  lune.  If  that  arc 
be  not  BC  let  it  be  BL,  an  arc  less  than  BC,  so  that 

surface  of  sphere  :  lune  ACEB  :  :  BZmDn  :  BL.  (1) 

Conceive  the  circumference  BCmDji  divided  into 
equal  parts,  each  of  which  is  less  than  CL,  the  as- 
sumed difi'erence  between  BC  and  BL.  Then  con- 
ceive one  of  these  equal  parts  applied  to  BC  as  a 
measure,  beginning  at  B.     Since  the  measure  is  les.<» 


OF  LUXES.  233 

than  LC,  one  point  of  division,  at  least,  will  fall  between  L  and  C.    Let  I  be 

such  a  point,  and  pass  the  arc  of  a  great  circle  through  A  and  I. 

Now,  surface  of  spJiere  :  lune  A\EB  ::  BCmDn  :  Bl,    (2) 

since  the  arc  Bl  is  commensurable  with  the  circumference.     In  (1)  and  (2),  the 

antecedents  being  equal,  the  consequents  should  be  proportional,  hence  we 

should  have 

lumkZEB   :  lune  A\EB   :  :  BL  :  Bl. 

But  this  is  absurd,  since  hine  ACEB  >   lune  AIEB,  whereas  BL  <  Bl.    In 
a  similar  manner  we  can  show  that 

surface  of  sphere  is  not  to  lune  ACEB  :  :  BCmDn  ;  any  arc  greater  than  BC. 

Hence,  as  the  fourth  term  can  neither  be  less  nor  greater  than  BC,  it  must 
be  equal  to  BC,  and  we  have 

surf  ace  of  sphere  :  lune  ACEB  ::  BCmD/i  :  BC, 
».  e.j  as  4  right  angles,  to  the  angle  of  the  lune. 

€10,  SCH,  2. — To  obtain  the  area  of  a  lune  whose  angle  is  knoicn,  on  a  c/iven 
ipliere,  find  the  area  of  the  sphere,  and  multiply  it  by  the  ratio  of  the  angle 
of  the  lune  (in  degrees)  to  360°.  Thus,  R  being  the  radius  of  the  sphere, 
^TtW  is  the  surface  of  the  sphere ;  and  the  lune  whose  angle  is  30°  is  -3^%  or 
•h  the  surface  of  the  sphere,  i.  e.,  -h  of  47rR^  =  i^rR^ 


PROPOSITION  XXIX. 

Oil,  Hieorein, — Tf  Uvo  semicircumferences  of  great  circles 
intersect  on  the  surface  of  a  hemisphere^  the  sum  of  the  two  opj^osife 
triangles  thus  formed  is  equivalent  to  a  lune  whose  angle  is  that 
included  hy  the  semicircumferences, 

Dem. — Let  the  semicircumferences  CEB  and 
DEA  intersect  at  E  on  the  surface  of  the  hemi- 
sphere whose  base  is  CABD  ;  then  the  sum  of  the 
triangles  CED  and  AEB  is  equivaleut  to  a  lune 
whose  angle  is  AEB. 

For,  let  tlie  semicircumferences  CEB  and  DEA 
rbe  produced  around  the  sphere,  intersecting  on 
the  opposite  hemisphere,  at  the  extremity  F  ot 
the  diameter  through  E.  Now,  FBEA  is  a  lune 
whose  angle  is  AEB.  Moreover,  the  triangle  AFB 
is  equivalent  to  the  triangle  DEC:    since  angle  ^ig  350. 

AFB  =  AEB  -  DEC,  side  AF  —  side  ED,  each  being 

the  supplement  of  AE;  and  BF  =:  CE,  each  being  the  supplement  of  EB. 
Hence,  tlie  sum  of  the  triangles  CED  and  AEB  is  equivalent  to  the  lune  FBEA. 
Q.  £.  D. 


234 


ELEMENTARY  SOLID   GEOMETRY. 


PROPOSITION  XXX. 

612,  Theorem, — The  area  of  a  spherical  triangle  is  to  the  area 
of  the  surface  of  the  hemisphere  i7i  which  it  is  situated,  as  its  spheri- 
cal excess  is  to  four  right  angles,  or  360°. 

Dem. — Let  ABC  be  a  spherical  triangle  wliose  angles  are  represented  by  A, 
B,  and  C ;  tlien  is 

area  ABC  :  surf,  of  hemisphere  :  :  A  +  B  +  C  —  180°  :  4  right  angles,  or  360°. 
Let  lune  A  represent  the  lune  whose  angle  is  the  an- 
gle A  of  the  triangle,  i.  e.,  angle  CAB,  and  in  like  man- 
ner understand  lune  B  and  lune  C. 

Now,  triangle  AHC+  AED   -  lune  A    {6H\ 
BHI  +  BEF   =  lune  B, 
CCF  +  CDI    =  lune  C. 
Adding,    2ABC  +  hemisphere  =  lune  (A  +  B  +  C)*,  (1) 
since  the  six  triangles  AHC,  AED,  BHI,  BEF,  CCF,  and 
CDI,   make    the  whole    hemisphere    and   2ABC    be- 
sides, ABC  being  reckoned  three  times.    From  (1),  we 
have  by  transposing  and  remembering  that  a  hemi- 
sphere is  a  lune  whose  angle  is  180°,  and  dividing 
by  2, 

ABC  =  ^lune  (A  +  B  +  C  -  180^). 
But,  by  {607\ 

i  lune  (A  +  B  +  C  -  180°) :  surf,  of  hemisph. :  :  A  +  B  +  C  -  180°  :  4  right  angles. 
Therefore,  ABC  :  surf,  of  liemisph.  :  :  A  +  B  +  C  —  180°  :  4  right  angles. 

613,  ScH.  \.—To  find  the  area  of  a  spherical  tiiangle  on  a  given  sphere, 

the  angles  of   the    triangle  being   given,  we  have  simply  to  multiply  the 

area  of  the   hemisphere,  i.  e.,  27rR',  by  the  ratio  of  the  spherical  excess 

to  360°.    Thus,  if  the  angles  are  A  =  110°,  B  =  80°,  and  C  =  50°,  we  have 

AD/-        O    TDS.       A+B  +  C-180°       „    ^3         60        ,^T„ 
area  ABC  =  27rR'»  x  ^— ^ =  2ii^  ^  S60  ~  ^  ^^  ' 

614L,  ScH.  2.— This  proposition  is  usually  stated  thus:  The  area  of  a 
spherical  triangle  is  equal  to  its  sph^?ical  excess  multiplied  by  the  trirectangular 
triangle.  When  so  stated  the  spherical  excess  is  to  be  estimated  in  terms  of 
the  right  angle ;  i.  e.,  having  subtracted  180°  from  the  sum  of  its  angles,  we  are 
to  divide  the  remainder  by  90°,  thus  getting  the  spherical  excess  in  right  angles. 
In  the  example  in  the  preceding  scholium,  the  spherical  excess  estimated  in  this 
110°  +  80°j^^50°-180°  ^  i.  and  the  area  of  the  triangle  would 


way  would  be 


90= 


*  This  signifies  the  lune  whose  angle  is  A  +  B  +  C,  which  ie  of  course  the  sum  of  tho 
three  lunes  whose  angles  are  Ai  B,  and  C- 


YOLUME    OF  SPHERE. 


235 


be  I  of  the  trirectangular  triangle.    Now,  the  trirectangular  triangle  being  i  of 
the  surface  of  the  sphere  {577)  is  i  of  47rR-',  or  ^nlC.     This  multiplied  by  | 
gives  ^7tW,  the  same  as  above. 
The  proportion, 

ABC  :  surf,  of  Tiemisph.  : :  A  +  B  +  C  —  180'  :  360°, 
is  readily  put  into  a  form  which  agrees  with  the  enunciation  as  given  in  this 
scholium.     Thus,  surf.ofhemisph.  =  27rR'',  whence 

A  +  B  +  C-180°       ,     ^,        A+B+C-ISO" 


ABC=r  2;rR'^  x- 


360= 


^7tW 


90° 


VOLUME  OF  SPHERE. 


PROPOSITION  XXXI. 

6 IS,  TJieorem. — The  volume  of  a  spliere  is  equal  to  the  area 
of  its  surface  multijflied  hy  \  of  the  raditis,  that  is,  JttR',  E  deing 
the  radius. 

Dem. — Let  OLrrR  be  the  radius  of  a  sphere. 
Conceive  a  circumscribed  cube,  that  is,  a  cube  whose 
faces  are  tangent  planes  to  the  sphere.  Draw  lines 
from  the  vertices  of  each  of  the  polyedral  angles  of 
the  cube,  to  the  centre  of  the  sphere,  as  BO,  CO,  DO, 
AO,  etc.  These  lines  are  the  edges  of  six  pyramids, 
having  for  their  bases  the  faces  of  the  cube,  and  for 
a  common  altitude  the  radius  of  tJie  sphere  (?). 
Hence  the  volume  of  the  circumscribed  cube  is 
equal  to  its  surface  multiplied  by  ^R. 

Again,  conceive  each  of  the  polyedral  angles  of 
the  cube  truncated  -by  planes  tangent  to  the  sphere.  A  new  circumscribed  solid 
will  thus  be  formed,  whose  volume  will  be  nearer  that  of  the  spbere  than  is  that 
of  the  circumscribed  cube.  Let  abc  represent  one  of  these  tangent  planes.  Draw 
from  the  polyedral  angles  of  this  new  solid,  lines  to  the  centre  of  the  sphere,  as 
aO,  50,  and  cO,  etc. ;  these  lines  will  form  the  edges  of  a  set  of  pyramids  whose 
"bases  constitute  the  surface  of  the  solid,  and  whose  common  altitude  is  the 
radius  of  the  sphere  (?).  Hence  the  volume  of  this  solid  is  equal  to  the  product 
of  its  surface  (the  sum  of  the  bases  of  the  pyramids)  into  ^R. 

Kow,  this  process  of  truncating  the  angles  by  tangent  planes  may  be  con- 
ceived as  continued  indefinitely ;  and,  to  whatever  extent  it  is  carried,  it  will 
always  be  true  that  the  volume  of  the  solid  is  equal  to  its  surface  multiplied  by 
iR.  Therefore,  as  the  sphere  is  the  limit  of  this  circumscribed  solid,  we  have 
the  volume  of  the  sphere  equal  to  the  surface  of  the  sphere,  which  ia  4;rR'', 
multiplied  by  ^R,  i.  e.,  to  ^3  7rR'.    q.  e.  d. 


Fig.  352. 


236  ELEMENTARY  SOLID  GEOMETRY. 

616.  Cor. — Tlie  surface  of  the  spliere  may  he  conceived  as  con- 
sisting of  an  infinite  numler  of  infinitely  small  plane  faces,  and  the 
volume  as  composed  of  an  infinite  number  of  jjyrajnids  having  these 
faces  for  their  bases,  and  their  vertices  at  the  centre  of  the  sphere,  the 
common  altitude  of  the  pyramids  being  the  radius  of  the  sphere. 

617 •  A  Spherical  Sector  is  a  portion  of  a  sphere  generated 
by  the  revohition  of  a  circular  sector  about  the  diameter  around 
which  the  semicircle  which  generates  the  sphere  is  conceived  to 
revolve.  It  has  a  zone  for  its  base ;  and  it  may  have  as  its  other  sur- 
faces one,  or  two,  conical  surfaces,  or  one  conical  and  one  plane 
surface. 

III. — Thus  let  db  be  the  diameter  around  which 
the  semicircle  aCh  revolves  to  generate  the  sphere. 
The  solid  generated  by  the  circular  sector  AOa  will 
be  a  spherical  sector  having  a  zone  (AB)  for  its  base  ; 
and  for  its  other  surface,  the  conical  surfoce  gene- 
rated by  AO.  The  spherical  sector  generated  by 
COD,  has  the  zone  generated  by  CD  for  its  base ;  and 
for  its  other  surfaces,  the  concave  conical  surface 
generated  by  DO,  and  the  convex  conical  surface 
generated  by  CO.  The  spherical  sector  generated 
by  EOF,  has  the  zone  generated  by  EF  for  its  base, 

the  plane  generated  by  EO  for  one  surface,  and  the  concave  conical  s-orface 

generated  by  FO  for  the  other. 

618.  A  Spherical  Seg^nent  is  a  portion  of  the  sphere  in- 
cluded by  two  parallel  planes,  it  being  understood  that  one  of  the 
planes  may  become  a  tangent  plane.  In  the  latter  case,  the  seg- 
ment has  but  one  base ;  in  other  cases,  it  has  two.  A  spherical 
segment  is  bounded  by  a  zone  and  one,  or  two,  plane  surfaces. 


PROPOSITION   XXXU. 

619.  Tlieorem. — The  volume  of  a  spherical  sector  is  equal  to 
the  pi'oduct  of  the  zone  which  forms  its  base  into  one-third  the  radius 
of  the  sphere. 

Dem. — A  spherical  sector,  like  the  sphere  itself,  may  be  conceived  as  con- 
sisting of  an  infinite  number  of  p\Tamids  whose  bases  make  up  its  surface,  and 
whose  common  altitude  is  the  radius  of  the  sphere.  Hence,  the  volume  of  the 
sector  is  equal  to  the  sum  of  the  bases  of  these  pyramids,  that  is,  the  spherical 
surface  of  the  sector,  multiplied  by  one-third  their  common  altitude,  which  is 
one-third  the  radius  of  the  sphere.    Q.  E.  D. 


EXERCISES  ON  THE  SPHERE. 


237 


620,  Cor. — The  volumes  of  spherical  sectors  of  the  same  or  equal 
spheres  are  to  each  other  as  the  zones  which  form  their  hases ;  and, 
since  these  zones  are  to  each  other  as  their  altitudes  (604),  the  sec- 
tors are  to  each  other  as  the  altitudes  of  the  zones  which  form  their 
bases. 


PROPOSITION  xxxm. 

621.  Theorein, — The  volume  of  a  spherical  segment  of  one  hase 
is  7rA'^(R  —  -JA),  A  leing  the  altitude  of  the  segment,  and  R  the  ra- 
dius of  the  sphere. 

Dem. — Let  CO  =  R,  and  CD  =  A ;  then  is  the  volume  of  the  spherical  seg- 
ment generated  by  the  revolution  of  CAD  about  CO 
equal  to  ;rA^(R  -  \L). 

For,  the  volume  of  the  spherical  sector  generated 
by  AOC  is  the  zone  generated  by  AC,  multiplied  by 
^R,  or  27rAR  x  ^R  =  ^ttAR^.  From  this  we  must 
subtract  the  cone,  the  radius  of  whose  base  is  AD,  and 
whose  altitude  is  DO.  To  obtain  this,  we  have  DO 
=  R  —  A :  whence,  from  the  right  angled  triaugle 
ADO,  AD  =  y'R^  -  (R  -  Ap"  =  v^2AR  -  A^  Now, 
the  volume  of  this  cone  is 

iQD  X  ttAD",  or  ^;r(R  -  A)  (2AR  -  A^)  =  ^7r(2AR2 
Subtracting  this  from  the  volume  of  the  spherical  sec- 
tor, we  have 

\TtKW  -  i;r(3AR2  -  SA^R  +  A^O  = 

^(A^R  -  lA')  =  TT A2(R  -  lA).    Q.  E.  D. 

Q22,  ScH.— The  volume  of  a  spherical  segment 
with  two  bases  is  readily  obtained  by  taking  the 
difference  between  two  segments  of  one  base  each. 
Thus,  to  obtain  the  volumes  of  the  segment  generated 
by  the  revolution  of  Z>CAc  about  «0,  take  the  differ- 
ence of  the  segments  whose  altitudes  are  ac  and  ah. 


Fig.  354. 


3A-^R  +  A^'). 


Fig.  355. 


EXERCISES. 

1.  What  is  the  circumference  of  a  small  circle  of  a  sjohere  whose 
diameter  is  10,  the  circle  being  at  3  from  the  centre? 

Ans.,  25.1328. 

2.  Construct  on  the  spherical  blackboard  a  spherical  angle  of  60°. 
Of  45°.    Of  90°.     Ofl20S     Of  250°. 


238  ELEMENTARY  SOLID  GEOMETRY. 

Sug's. — Let  P  be  the  point  where  the  vertex  of  the  required  angle  is  to  be 
situated.  With  a  quadrant  strike  an  arc  from  P,  wliich  shall  represent  one  side 
of  the  required  angle.  From  P  as  a  pole,  with  a  quadrant,  strike  an  arc  from 
the  side  before  drawn,  which  shall  measure  the  required  angle.  On  this  last  arc 
lay  off  from  the  first  side  the  measure  of  the  required  angle,*  as  60°,  45°,  etc. 
Through  the  extremity  of  this  arc  and  P  pass  a  great  circle  {548).  [The  stu- 
dent should  not  fiiil  to  give  the  reasons,  as  well  as  do  the  work.] 

3.  On  the  spherical  blackboard  construct  a  spherical  triangle  ABC, 
having  ab  =  100°,  AC  =  80°,  and  A  =  58°. 

4.  Construct  as  above  a  spherical  triangle  ABC,  having  AB  =  75°, 
A  =  110°,  and  B  =  87°. 

5.  Construct  as  above,  having  AB  =  150°,  BC  =  80°,  and  AC  =  100°- 
Also  having  AB  =  160%  AC  =  50°,  and  BC  =  85^ 

6.  Construct  as  above,  having  A  =  52°,  AC  =  47°,  and  CB  =  40°. 

Sug's. — Construct  the  angle  A  as  before  taught,  and  lay  off  AC  from  A  equal 
to  47°,  with  the  tape.  This  determines  the  vertex  C.  From  C,  as  a  pole,  with 
an  arc  of  40°,  describe  an  arc  of  a  small  circle ;  in  this  case  this  arc  will  cut  the 
opposite  side  of  the  angle  A  in  two  places.  Call  these  points  B  and  B'.  Pass 
circumferences  of  great  circles  through  C,  and  B,  and  B'.  There  are  two  tri- 
angles, ACB  and  ACB'. 

Note. — The  teacher  can  multiply  examples  like  the  three  preceding  at  pleas- 
ure. This  exercise  should  be  continued  till  the  pupil  can  draw  a  spherical  tri- 
angle as  readily  as  a  plane  triangle. 

7.  "What  is  the  area  of  a  spherical  triangle  on  the  surface  of  a 
sphere  whose  radius  is  10,  the  angles  of  the  triangle  being  85°, 
120°,  and  150°?  Aiis.,   305.4  +, 

8.  What  is  the  area  of  a  spherical  trfkngle  on  a  sphere  whose 
diameter  is  12,  the  angles  of  the  triangle  being  82°,  98°,  and  100°  ? 

9.  A  sphere  is  cut  by  5  parallel  planes  at  7  from  each  other.  What 
are  the  relative  areas  of  the  zones  ?     What  of  the  segments  ? 

10.  Considering  the  earth  as  a  sphere,  its  radius  would  be  3958 
miles,  and  the  altitudes  of  the  zones,  North  torrid  =  1578,  North 
temperate  =  2052,  and  North  frigid  =  328  miles.  What  are  the 
relative  areas  of  the  several  zones  ? 

SuG. — The  student  should  be  careful  to  discriminate  between  the  width  of  a 
zone,  and  its  altitude.  The  altitudes  are  found  from  their  widths,  as  usually 
given  in  degrees,  by  means  of  trigonometry. 

*  For  this  purpose  a  tape  equal  in  length  to  a  semicircumference  of  a  great  circle  of  the 
»phere  used,  and  marked  off  into  180  equal  parts,  will  be  convenient.  A  strip  of  paper  ma^ 
be  need. 


EXERCISES  ON  THE  SPHERE.  239 

11.  The  earth  being  regarded  as  a  sphere  whose  radius  is  3958 
miles,  what  is  the  area  of  a  spherical  triangle  on  its  surface,  the 
angles  being  120°,  130°,  and  150°  ?  What  is  the  area  of  a  trirectan- 
gular  triangle  on  the  earth's  surface  ? 

12.  Construct  on  the  spherical  blackboard  a  spherical  triangle 
ABC,  having  A  =  59°,  AC  =  120°,  and  AB  =  88°.  Then  construct  the 
triangle  polar  to  ABC. 

13.  Construct  triangles  polar  to  each  of  those  in  Examples  3,  4, 
and  5. 

14.  In  the  spherical  triangle  ABC  given  A  =  58°,  B  =  67°,  and 
AC  =  81° ;  what  can  you  affirm  of  the  polar  triangle  ? 

15.  ■  What  is  the  volume  of  a  globe  which  is  2  feet  in  diameter  ?  What 
of  a  segment  of  the  same  globe  included  by  two  parallel  planes,  one 
at  3  and  the  other  at  9  inches  from  the  centre  ? 

16.  Compare  the  convex  surfaces  of  a  sphere  and  its  circumscribed 
cylinder  and  cone,  the  vertical  angle  of  the  cone  being  60°. 

17.  Compare  the  volumes  of  a  sphere  and  its  circumscribed  cube, 
cylinder,  and  cone,  the  vertical  angle  of  the  cone  being  60°. 

18.  If  a  and  l  represent  the  distances  from  the  centre  of  a  sphere 
whose  radius  is  r,  to  the  bases  of  a  spherical  segment,  show  that  the 
volume  of  the  segment  is  ;r[r'  (b  —  a)  —  \{¥  —  a')].  See  (62 1, 
622). 


^Y3 


PART  III. 

AN    ADVANCED   COURSE 
IN     GEOMETRY. 


OHAPTEE  I. 

EXERCISES    IX    GEOMETRICAL    INVENTION, 


SECTION  I. 

THEOREMS  IN  SPECIAL  OR  ELEMENTARY  GEOMETRY. 

S2S.  Tliis  chapter  will  afford  a  review  of  Parts  I.  and  II.,  while 
it  will  greatly  extend  the  student's  knowledge  of  geometrical  facts. 
Great  pains  should  be  taken  to  secure  good  habits  as  to  neatness  of 
execution  in  the  construction  of  figures,  orderly  and  proper  arrange- 
ment of  thought,  and  in  style  of  expression.  The  practice  of  con- 
structing every  figure  upon  geometrical  principles — guessing  at 
nothing — cannot  be  too  strongly  commended.  As  to  the  form  of  a 
geometrical  argument,  observe  the  following  order : 

1st.  The  enunciation  of  the  theorem  or  problem  in  general  terms. 

2d.  The  elucidation  of  the  general  statement,  by  reference  to  the 
particular  figure  which  it  is  proposed  to  use. 

3d.  A  description  of  the  figure,  with  reference  to  any  auxiliary 
construction  which  is  used  in  the  demonstration  or  solution. 

4th.  The  demonstration  proper. 


S24z.  If  two  adjacent  sides  of  a  quadrilateral  are  equal  each  to 
each,  and  the  other  two  adjacent  sides  equal  each  to  each,  the  diago- 
nals intersect  at  right  angles. 

SuG's.^st.  Draw  a  quadrilateral  having  such  sides  as  the  data  require,  and 
draw  its  diagonals.    2d.  State   the   proposition  witli  reference  to  the  figure. 


2U 


EXERCISES   IN   GEOMETEICAL  INVENTION. 


3d.  [In  this  case  the  regular  third  step  is  not  required,  as  no  auxiliaiy  lines  are 
necessarj',]  4th.  Prove  that  the  diagonals  are  at  right  angles  to  each  other. 
The  demonstration  is  based  upon  a  corollary  in  Section  I.,  Part  II.,  Chapter  I. 

62o.  Cor. — One  of  the  diagonals  is  bisected.     [State  which  one, 
and  show  why.] 


626.  If  a  parallelogram  has  one  oblique  angle,  all  its  angles  are 
oblique ;  and  if  it  has  one  right  angle,  all  its  angles  are  right  angles. 

Sug's. — Let  the  student  be  careful  to  follow  the  order  as  heretofore  given. 
No  auxiliary  construction  is  needed.  The  demonstration  is  based  upon  the 
doctiine  of  parallels. 


627'  The  sum  of  three  straight  lines  drawn  from  any  point 
within  a  triangle  to  the  vertices  is  less  than  the  sum,,  and  greater 
thau  the  half  sum  of  the  three  sides  of  the  triangle. 

Sug's. — The  first  statement  is  proved  from  (270)  and  the  second  from  {274:.) 


628.  A  line  drawn  from  any  angle  of  a  triangle  to  the  middle 
of  tlie  opposite  side,  is  less  than  the  half  sum  of  the 
adjacent  sides,  and  greater  than  the  difference  between 
this  half  sum  and  half  the  third  side. 


Sug's. — 1st.  Draw  a  triangle,  as  ABC,  bisect  one  side,  as  AC, 
and  draw  BD.  2d.  Make  the  statement  with  reference  to  the 
figure.  3d.  Produce  BD  until  DE  =  BD,  and  draw  AE  and  EC. 
4th.  The  first  step  in  the  proof  is  to  show  the  triangle  ADE  equal 
to  CBD,  and  ADB  equal  to  DCE ;  whence  AE  =  BC,  and  EC  =  AB. 


Fig.  356. 


Fig.  357. 


629.   If   hues   be   drawn  from  the  extremities  of 
either  of  the  non-parallel  sides  of  a  trapezoid  to  the  mid- 
dle  of   the    opposite   side,    the  triangle  thus 
formed  is  half  the  trapezoid. 

StjG's. — Tlie  third  step,  or  construction,  consists  in 
drawing  HE  parallel  to  AD  and  hence  to  BC  (?),  and 
FG  through  E  parallel  to  AB. 


630.  Any  line  drawn  through  the  centre  of  the  diagonal  of  a 
parallelogram  bisects  the  figure. 


THEOREMS   IN   SPECIAL   GEOMETRY. 


245 


031.  Prove  that  the  sum  of  the  angles  of  a 
triangle  is  two  right  angles,  by  producing  two 
of  the  sides  about  an  angle  and  through  this 
angle  drawing  a  line  parallel  to  the  third  side. 

Prove  the  same  by  producing  one  side  of  the 
triangle  and  drawing  a  line  through  the  ex- 
terior angle  parallel  to  the  non-adjacent  side. 


Fig.  3o8. 


632.  If  any  point,  not  the  centre,  be  taken 
in  a  diameter  of  a  circle,  of  all  the  chords 
which  can  pass  through  that  point,  that  one  is 
the  least  w^hich  is  at  right  angles  to  the  diameter. 

®  

033.  If  from  any  point  there  extend  two  lines  tangent  to  a  cir- 
cumference, the  angle  contained  by  the  tangents  is  double  the  angle 
contained  by  the  line  joining  the  points  of  tangency  and  the  radius 
extending  to  one  of  them. 


034:.  The  angle  included  by  two  lines  drawn  from  any  angle  of  a 
triangle,  the  one  bisecting  the  angle  and  the 
other  perpendicular  to  the  opposite  side,  is 
half  the  difference  of  the   other  two  angles 
of  the  triangle. 


Sug's.  ABD  =  90°  -  A,  whence  ABD  -  EBD  = 
90°  -  A  -  EBD.     Also,  DBC   =  90°  -  C,  whence  EBC  =  90^ 
.-=  90°  -  A  -  EBD.  etc. 


C  +  EBD 


035.  If  three  lines  be  drawn  from  the 
angled  triangle — two  bisecting  these  angles, 
and  a  third  a  perpendicular  to  one  of  the 
bisecting  hues — the  triangle  included  by 
these  lines  will  be  isosceles. 


acute  angles 


of  a 


riijht 


"Sug's. — It   is   to  be    proved  that  OD 
COD  =  OAC  +  ACQ  =  45°,  etc. 


CD. 


030,  If  one  circumference  be  described 
on  the   radius  of  another  as   a  diameter, 

any  straight  line  extending  from  their  point  of  contact  to  the  outer 
circumference  is  bisected  by  the  inner. 

SuG.— The  demonstration  is  based  upon  {159,  211). 


246 


EXERCISES  IN   GEOMETRICAL  INVENTION. 


637 •  Prove  that  the  sum  of  the  angles  of  a  regular  five  point  star 
(Fig.  101)  is  two  right  angles.  Show,  also,  that  the  figure  formed  by 
the  iuterceiDted  portions  of  the  lines  is  a  regular  pentagon. 


038.  If  the  sides  of  a  regular  hexagon  are  produced  till  they 
meet,  show  that  the  exterior  figures  will  be  equilateral  triangles. 

639»  If  from  two  given  points  on  the  same 
side  of  a  given  line,  two  lines  be  drawn  meet- 
ing in  the  line,  their  sum  is  least  when  they 
make  equal  angles  with  the  line. 


Fig.  361. 


640.  If  from  tT\'o  given  points  without  a 
circumference,  two  lines  be  drawn  meeting  in  the  circumference, 
their  sum  is  least  when  they  make  equal  angles  with  a  tangent  at 
the  common  point,  the  points  being  on  the  opposite  side  of  the  tan- 


gent from  the  circle. 


641.  The  side  of  an  equilateral  triangle  inscribed-  in  a  circle  is 
equal  to  the  diagonal  of  a  rhombus,  whose  other  diagonal  and  each 
of  whose  sides  are  equal  to  the  radius. 


642.  If  two  circumferences  intersect  each  other,  and  from  either 
point  of  intersection  a  diameter  be  drawn  in  each,  the  other  ex- 
tremities of  these  diameters  and  the  other  point  of  intersection  are 
in  the  same  straight  line. 


643.  If  any  straight  line  joining  two  parallels  be  bisected,  any 
other  line  through  the  point  of  bisection  and  included  by  the  par- 
allels, is  bisected  at  the  same  point. 


644.  If  the  sides  of  any  quadrilateral  are  bisected,  the  quadri- 
lateral formed  by  joining  the  adjacent  points  of  bisection  is  a  par- 
allelogram. 

SuG*s. — 1st.  Draw  a  quadrilateral,  bisect  its 
sides,  and  join  the  adjacent  points  of  bisection. 
2d.  State  the  proposition,  with  reference  to  the 
figure.  3d.  Draw  the  diagonals.  4lh.  Give  the 
proof.    It  is  based  on  the  similarity  of  triangles. 

645.  Cor.  L— The  parallelogram  is  one- 
half  the  trapezium.     Prove  it.     What  figure 
is  formed  by  joining  the  centres  of  EF,  FC, 
Fig.  ^n.  and  FC,  HG,  etc.  ? 


THEOREMS  IN   SPECIAL   GEOMETRY.  247 

64G.  Cor.  2. — Lines  joining  the  middle  points  of  the  opposite 
sides  of  any  trapezium  bisect  each  other  (?). 


6^7'  If  two  straight  lines  join  the  alternate  ends  of  two  parallels, 
the  line  joining  their  centres  is  half  the  dif- 
ference of  the  parallels.  a,^ ^- 

Sug's.— We  are  to  prove  that  EF  =  i  (CD  —  AB). 
CH  =  EF  =  i  (CD  -  AB). 


64:8.  In  any  right-angled  triangle  the  line  drawn  from  the  right 
angle  to  the  middle  of  the  hypotenuse  is  equal  to  one-half  the 
hypotenuse. 

04:9.  The  perpendiculars  which  bisect  the  three  sides  of  a  triangle 
meet  in  a  common  point. 

Sug's. — Eirst  show  that  the  intersection  of  two  of  the  perpendiculars  is 
equally  distant  from  the  three  vertices  of  the  triangle.  Then  that  a  line 
drawn  from  this 'point  to  the  middle  of  the  third  side  is  perpendicular  to  it. 


650.  The  three  perpendiculars  drawn  from  the  angles  of  a  tri 
angle  upon  the  opposite  sides  intersect 
in  a  common  point. 


■;N 


Sug's. — Draw  through  the  vertices  of  the 
triangle  lines  parallel  to  the  opposite  sides. 
The  proposition  may  then  be  brought  under 
the  preceding. 

651.  Cor. — The  following  triangles  ''•../ 

are   similar — viz.,  BOE,  BDC,  AOD,  and  w 

'aec,  each  to  each  ;  also  BOF,  BDA,  DOC, 

and  CFA.     Prove  it. 


Fig.  364. 


652.  If  from  a  point  without  a  circle  two  secants  be  drawn,  mak- 
ing equal  angles  with  a  third  secant  passing  through  the  same  point 
and  the  centre  of  the  eircle,  the  intercepted  chords  of  the  lirst  two 
are  equal. 

SuG.  --Prove  by  revolving  one  part  of  the  figure. 


248 


EXERCISES   IN   GEOMETRICAL  INVENTION. 


6ij3.  The  Slim  of  the  alternate  angles  of  any  hexagon  inscribed 

in  a  circle  is  four  riffht  ano:les. 


634:.  If  two  circles  intersect  in  A  and  B, 
and  from  P,  any  point  in  one  circumference, 
the  chords  PA  and  PB  be  drawn  to  cut  the 
other  in  c  and  D,  CD  is  parallel  to  a  tangent 
at  P. 


Fig.  365. 

pendicular  to  each  other. 


Goi>.  If  two  lines  intersect,  two  lines 
which  bisect  the  opposite  angles  are  per- 


656.  The  angle  included  by  two  lines  drawn  from  a  point  within 
a  triangle  to  the  vertices  of  two  of  the  angles,  is  greater  than  the  third 
angle. 

Sug's.— The  demonstration  may  be  founded  on  {219)  or  {231). 


657.  In  a  triangle  whose  angles  are  90°,  60°,  and  30°,  show  that 
the  longest  side  is  twice  the  shortest. 


658.  Lines  which  bisect  the  adjacent  angles  of  a  parallelogram 
are  mutually  pei-pendicular. 


659.  If  from  any  point  in  the  base  of  an  isosceles  triangle  lines 
are  drawn  parallel  to  the  sides,  a  parallelogram  is  formed  whose  peri- 
meter is  constant  and  equal  to  the  sum  of  the  two  equal  sides  of  the 
triangle. 


Fio.  3G6. 


660.  If  from  any  point  in  the  base  of  an  isosceles 
triangle  perpendiculars  be  drawn  to  the  sides  of  the 
triangle,  their  sum  is  constant  and  equal  to  the  per- 
pendicular from  one  of  the  equal  angles  of  the  triangle 
upon  tlie  opposite  side. 


661.  If  from  any  poin«  within  an  equilateral  tri- 
angle, throe  perpondiculL  s  be  let  fall  upon  the  sides, 
their  sum  is  constant  and  equal  to  the  altitude  of  the 


Fig.  367. 


triangle. 


THEOREMS   IN   SPECIAL  GEOMETRY. 


0^ 

602.  If  from  a  fixed  point  without  a  circle  two 
tangents  be  drawn  terminating  in  the  circumfer- 
ence, the  triangle  formed  by  them  and  any  tan- 
gent to  the  included  arc  has  a  constant  perimeter 
equal  to  the  sum  of  the  first  two  tangents. 


249 


Fig.  3fi8. 


063,  The  sum  of  two  opposite  sides  of  a  quadrilateral  circum- 
scribed about  a  circle,  is  equal  to  the  sum  of  the  other  two. 


004.  If  two  opposite  angles  of  a  quadrilateral  are  supplemen- 
tary, it  may  be  circumscribed  by  a  circumference. 


005.  The  square  described  on  the  sum  of  two 
lines  is  equivalent  to  the  sum  of  the  squares  on 
the  lines,  ^;Z?^5  twice  the  rectangle  of  the  lines. 

Sug'r. — Be  careful  to  give  the  construction  fully,  and 
show  that  the  parts  are  rectangles,  etc. 


a     b  6' 

"*  a  b 


Fig.  309. 


000.  The  square  described  on  the  difference  of 
two  lines  is  equivalent  to  the  sum  of  tlie  squares 
on  the  lines,  minus  twice  the  rectangle  of  the 
lines. 


a-b 


007.  The  rectangle  of  the  sum  and  difference 
of  two  lines  is  equivalent  to  the  difierence  of  the 
squares  described  on  the  lines. 

Sen.— The  three  preceding  propositions  are  but  geo- 
metrical conceptions  and  demonstrations  of  the  algebraic 
farmul(B,  {a  +  hf  =  a"  +  2ab  +  b\  {a  -  bf  =  «'  -  2ab 
+  b\  and  {a  +  b)  {a  -  b)  ■=  «*  -  b\ 


r 

b' 

B 
3 

a-b 

b 

ft 

Fig.  371. 


250 


EXERCISES  IN  GEOMETRICAL  INTENTION. 


VARIOUS  DEMONSTRATIONS  OF  THE  PYTHAGOREAN  PROPOSITION. 

668.  The  square  described  on  the  liypotenuse  of  a  right-angled 
triangle  is  equivalent  to  the  sum  of  tiie  squares 
described  on  the  other  two  sides. 

1st  Method. — Let  ABC  be  the  given  trianiilo,  and 
ACED  the  square  described  on  the  hypotenuse.  Complete 
the  construction.  Show  that  the  four  triangles  are  equal. 
The  square  HF  is  (AB  -  BC)^  The  student  can  complete 
the  demonstration. 

Fig.  372. 


2d  IVIethgd. — Let  ACED  be  the  square  on  the 
hypotenuse.  Let  fall  the  perpendiculars  EF,  DC, 
etc.  Show  that  the  three  triangles  are  equal,  and 
that  FD  and  LB  are  the  squares  of  the  two  sides  AB 
and  BC. 


3d  Method.— Let  BL  and  BH  be  the 
squares  on  the  sides.  Produce  FL  and 
HG  till  they  meet  in  K.  Draw  DA  and  EC 
perpendicular  to  AC,  and  draw  DE  and 
KB.  Prove  that  ACED  is  a  square,  and  also 
that  the  triangles  ABC,  CLE,  BFK,  KBC, 
DKE,  and  AHD  are  all  equal  to  each  other. 
The  demonsti*ation  is  then  readily  made. 


4th  Method.— This  is  the  demonstration 
usually  given  in  our  text-books.  Drawing  the 
squares  on  the  three  sides,  let  fall  Bl  y>erpen- 
dicular  to  AC  and  produce  it  to  K.  Draw  BD, 
BE,  HC  and  AF.  Show  that  the  triangle  HAC 
=  BAD,  and  that  the  former  is  half  the  square 
AG,  and  the  latter  half  the  rectangle  AK. 
Hence  AG  =  AK.  In  like  manner  show  that 
LC  ==  CK. 


THEOREMS   IN   SPECIAL  GEOMETRY. 


251 


We  will  now  give  a  few  other  figures  by  means  of  which  the  demonstration 
can  be  effected,  and  leave  the  student  to  his  own  resources  in  effecting  it. 


r.^ ^ 


5th  Method. 


6th  Method.         7th  Method.  8th  "Method. 

Fig.  376. 


'  9th  Method. — The  truth  of  the  theorem  appears  also 
as  a  direct  consequence  of  {360). 


U" 


Fig.  377. 


669'  In  an  oblique  angled  triangle  the  square 
of  a  side  opposite  an  acute  angle  is  equivalent  to  the  sum  of  the 
squares  of  the  other  two  sides  diminished  by  twice  the  rectangle  of 
the  base,  and  the  distance  from  the  acute  angle  to  the  foot  of  the 
perpendicular  let  fall  upon  the  base  from  the  angle  opposite. 

SuG's.— It  is  to  be  shown  that  AB^  =  BC'  +  AC^  -  2AC  x  DC. 
Observe  that  AD'  =  (AC  -  DO'  =  AC  +  DC'  -  2AC  x  DC. 
Whence,  by  a  simple  application  of  the  preceding  theorem, 
the  truth  of  this  becomes  apparent. 


670'  In  an  obtuse  angled  triangle  the  square  of  the  side  opposite 
the  obtuse  angle  is  equivalent  to  the  sum  of  the  squares  on  the  other 
two  sides,  increased  by  twice  the  rectangle  contained  by  the  base  and 
the  distance  from  the  obtuse  angle  to  the  foot  of  the  perpendicular 
let  fall  from  the  augle  opposite  upon  the  base  produced. 

SuG. — The  demonstnition  is  analogous  to  the  preceding,  C  being  made  obtuse 
in  this  case ;  whence  AD  =  AC  +  DC,  etc. 


252 


EXERCISES  IN  GEOMETRICAL  INVENTION. 


671.  The  following  is  an  ontline  of  a  general  demonstration  cov- 
ering the  three  preceding  propositions: 

Letting  AE,  BD,  and  CF  be  the  three  perpendiculars  from 
the 'angles  upon  the  opposite  sides,  and  observing  that  a 
circumference  described  on  any  side  as  a  diameter  passes 
through  the  feet  of  two  of  the  perpendiculars,  {350)  and 
{355)  readily  give  the  following  : 

AB  X  AF  =^  AC  X  AD  rr  AC'  ±  AC  X  CD, 
and  AB  x  BF  =  BC  x  BE  ^  BC"  ±  BC  x  CE  j^ 

adding.  AB^  =  AC=  +  BC'  ±  2AC  x  CD  (or  2BC  x  CE), 
the  +  sign  being  taken  when  C  is  obtuse,  and  the  —  sign 
when  C  is  acute.   If  C  is  right  CE  and  CD  become  0,  whence  AB"'  =  AC'^  +  BC*. 


672.  Def. — The  line  drawn  from  any  angle  of  a  triangle  to  the 
middle  of  the  opposite  side  is  called  a  medial  line. 


673'  The  sum  of  the  sqnares  of  any  two  sides  of  a  triangle  is 
equivalent  to  twice  the  square  of  the  medial  line  drawn  from  their 
included  angle,  plus  twice  the  square  of  half  the  third  side. 

SuG.— Proved  by  applying  {669,  670). 


674.  The  three  medial  lines  of  a  triangle  mutually  trisect  each 
other,  and  hence  intersect  in  a  common  point. 

Sug's.— To  prove  that  OE  =  -^BE,  draw  FC  parallel  to 
AD  until  it  meets  BE  produced.  Then  the  triangles 
AEO  and  FEC  are  equal  (?);  whence  EF  =  OE.  Also, 
BO  =  OF  (?). 

Having  shown  that  OE  =  ^BE,  by  a  similar  construc- 
tion we  can  show  that  OD  =  ^AD- 

Finally,  we  may  show  that  the  medial  line  from  C  to  AB  cuts  off  ^  of  BE,  and 
hence  cuts  BE  at  the  same  point  as  does  AD- 

Another  Dem. — Lines  through  O  parallel  to  the  sides  trisect  tlie  sides,  etc 


67S.  In  any  quadrilateral  the  sum  of  the  squares  of  the  sides  is 
equivalent  to  the  sum  of  the  squares  of  the  diagon- 
als, plus  four  times  the  square  of  the  line  joining  the 
centres  of  the  diagonals. 

676.  Cor.— The  sum  of  the  squares  of  the  sides 
of  a  parallelogram  is  equivalent  to  the  sum  of  the 
Fig.  osi.  squares  of  the  diagonals. 


THEOREMS   IN   SPECIAL   GEOMETRY.  253 

677.  In  any  quadrilateral  which  may  be  inscribed  in  a  circle, 
the  product  of  the  diagonals  is  equal  to  the  sum  of  the  products  of 
the  opposite  sides. 

678.  In  any  triangle  the  rectangle  of  two  sides  is  equivalent  to 
the  rectangle  of  the  perpendicular  let  fall  from  their 

included  angle  upon  the  third  side,  into  the  diameter 
of  the  circumscribed  circle. 

SuG. — This  proposition  is  an  immediate  consequence  of  the 

similarity  of  two  triangles  in  the  figure. 

Fig.  382. 

670.  Cor. — The  area  of  a  triangle  is  equivalent  to  the  product 
of  its  sides  divided  by  twice  the  diameter  of  the  circumscribed  circle. 


680.  If  there  be  an  isosceles  and  an  equilateral  triangle  on  the 
same  base,  and  if  the  vertex  of  the  inner  triangle  is  equally  dis- 
tant from  the  vertex  of  the  outer  one  and  from  the  ends  of  the  base, 
then,  according  as  the  isosceles  triangle  is  the  inner  or  the  outer 
one,  its  base  angle  will  be  J  of,  or  2J  times  the  vertical  angle. 


681,  Of  all  triangles  on  the  same  base,  and  having  the  same 

vertical  angle,  the  isosceles  has  the  greatest  area. 

Sug's. — Describe  a  segment  on  the  given  base,  which  shall  contain  the  given 
angle.  The  triangle  on  this  base  and  having  its  vertex  in  the  arc  of  the  seg- 
ment is  the  triangle  to  be  considered. 


682.  Two  triangles  are  similar,  when  two  sides  of  one  are  pro- 
portional to  two  sides  of  the  other,  and  the  angle  opposite  to  that 
side  which  is  equal  to  or  greater  than  the  other  given  side  in  one,  is 
equal  to  the  homologous  angle  in  the  other. 


ALGEBRAIC   DEMONSTRATIONS. 

683.  The  difference  of  the  squares  on  any  side  of  a  regular 
pentagon  and  any  side  of  regular  decagon,  inscribed  in  the  same 
circle,  is  equivalent  to  the  square  of  the  radius. 

Sug's.— We  will  give  the  outline  of  what  may  be  termed  an  Algebraic  Demon- 
stmtion  of  this  proposition.    This  method  is  often  the  most  convenient  and  ex- 


254  EXERCISES  IN  GEOMETRICAL  INVENTION. 

peditious.    Letting  p  represent  a  side  of  the  pentagon,  d  a  side  of  the  decagon, 
and  r  the  radius,  the  student  should  be  able  to  discover  the  following  relations : 

(1)  r'.d::d:r-  d.ovr^  -  dr  =  d^ ;    Xju^  <'^^  ^  ^*- 

(2)  Vd^  -  ip*  +  v/r*  -  ^'  =  ^•• 

From  (2),  2rVr^  —  ^p^  =  2r^  —  d*  =  r^  +  dr,  by  substituting  for  <f  its  value 
from  (1).  Hence  4r^  —  p^  =  r*  +  2dr  +  d^,  or  3r*  —  2dr  =  pi  +  d*.  In  this, 
substituting  the  value  of  dr  as  found  in  (1),  we  readily  obtain  r^  z=z  p^  —  <f. 

Q.   E.   D. 


084.  Demonstrate  algehraically  that  the  square  on  the  sum  of 
two  lines,  together  with  the  square  on  the  difference,  is  double  the 
sum  of  the  squares  on  the  lines  separatel}'. 


GSo'  The  sum  of  the  squares  of  the  three  medial  lines  of  a  tri- 
angle is  three-fourths  of  the  sum  of  the  squares  of  the  sides. 


686.  The  square  of  any  side  of  a  triangle  is  equivalent  to  twice 
the  sum  of  the  squares  of  the  segments  of  the  medial  lines  adjacent 
to  its  extremities,  minus  the  square  of  the  non-adjacent  segment  of 
the  third  medial  line. 

Deduced  algebraically  from  the  preceding. 


687.  The  sum  of  the  squares  of  the  three  greater  segments  of 
the  medial  lines  of  a  triangle  is  equivalent  to  one-third  the  sum  of 
the  squares  of  the  sides  of  a  triangle.     ^ 

Deduced  algebiaically  from  the  preceding. 


688.  The  lines  from  the  vertices  of  a  triangle  to  the  points  of 
tangency  of  the  inscribed  circle  intersect  in  a  common  point. 

SuG's.  DC  is  parallel  to  AF,  BD  =  BF  =  h,  CF  =  CP 

dr 

=  c,  AD  =  AP  =  «,  DC  =  d,   FC  =  e.       OF  =  -^^, 

c  +  e 

,       AF  X  6               ab  r^r-       AC  ^ 

d  =  -,  e  =  r.     .-.  OF  =  AF 


a  +  b'  a  +b'     "  ■        a{b  +  c)  +  be 

B      In  like  manner  we  may  find  where  PB  intersects  AF,  by 

drawing  through  p  a  parallel  to  AF.     This  distance  is 

Fia.  ZSi.  be 

found  to  be  OF  =  AF  X  — ^ -,  a  result  which 

a  (b  +  c)  +  be 

might  have  been  anticipated,  since  b  and  c  are  similarly  involved. 


THEOREMS  IN  SPECIAL  GEOMETRY.  255 

689.  The  area  of  a  triangle,  as  expressed  in  terms  of  its  sides,  is 
Area  =  the  square  root  of  the  continued  product  of  half  the  sum  of 
the  sides  into  this  half  surn  rninus  each  side  separately. 


Sug's,— We  will  give  the  outlines  of  both  the  Geometric  and  Algebraic  de- 
monstrations : 

1st.  Geometric  Demonstration.    CD=:CB,  CE  =  CA,  and  through  F,  the 
centre  of  DA,  HG  is  drawn  parallel  to  AB,  With  F  as  a  centre, 
and  FH  as  a  radius,  a  circumference  passes  through  G  (?).    ON  ^ 

is  perpendicular  to  AE  and  passes  through  H  (?).  i/'  \ 

Now  OF  =  i  (AC  +  CB),  and  FL  =  MB  (-).  /"    ^^^^^^^ 

Heuce  CL  =  \{kZ  +  CB  +  AB)  =  iS,  S  being  the  sum  of    \  J'jy^.....}.\^ 
th  e  sides ;  ^-'-Jt'^P  ^ 

Hence,  also,  DL  ==  Al  =  ^S  -CB,  CI  =  ^S  -  AB,  and  AL  -  i^S 

_  A  Q  "  Fig-  384. 

Again,  CN  x  AN  =  area  ACE ; 
and  HN  X  AN  =  area  ABE; 
Subtracting,  AN  (CN  -  HN)  =  AN  x  CH  =  area  ACB    (1). 
Once  more,  CH  x  DH  =  area  CDB; 
and  HN  X  DH  =  area  ADB. 
Adding,  DH  (CH  +  HN)  =  CA  x  CN  r=  area  ACB    (2). 
Multiplying  (1)  and  (2),  we  have 
GA  x  AN  X  CH  X  CN  =  (area  ACB)2. 
But  CN  X  CH  =  CL  X  CI,  and  GA  x  AN  =  AL  x  Al  =r  AL  x  DL. 
Therefore,  area  ACB  =:\/CL  x  CI  x  AL  x  DL  = 
v^iSTiS  -  AB)  (IS  -  AC)  as  -  CB). 


2d.  Algebraic  Demonstration.    From  the  right  angled  triangles  BCD  and 

ACD,vfe  find  m  = ^-^—. 

2c 

Whence  p  =  |/7r^(^IZ|I±Zy ,  and 

area  ABC  =  |  ^7^^^-?y  ,^^^ 

=  W-  a*  +  2a^6'^~2aV  -^*  +2/>V  -  c* 

=  i  V{a  +  b  +  c)  {a  +  b  -  c)  {a  -  b  +  c)  {-  a  +  b  +  c) 

=  v/i«(i{j  -  c)  (is  -  b)  (i«  -  a). 


A  D 


256 


EXERCISES  IN  GEOMETRICAL  INVENTION. 


090.  From  auy  poiut  in  the  plane  of  a  circle  the  greatest  and 
least  distances  to  the  circumference  are  measured  on 
the  line  passing  through  the  centre. 


Sug's. — There  are  three  cases  :— 1st.  When  the  point  is  with- 
out the  circle.  2d.  When  the  point  is  within.  3d.  When  the 
point  is  in  the  chcumference. 


Fie.  386. 


091'  From  any  point  except  the  centre  of  a  circle, 
two.  and  only  two,  equal  lines  can  be  drawn  to  the  cir- 
cumference. 

SuG. — This  is  a  direct  consequence  of  {181, 182). 


092.  If  two  opposite  sides  of  a  parallelogram  be  bisected,  straight 
lines  from  the  points  of  bisection  to  the  opposite  vertices  will  trisect 


the  diagonal. 


093.  The  feet  of  two  perpendiculars  let  fall  from  two  given  points 
upon  a  given  line  are  equally  distant  from  the  middle  of  the  line 
joining  the  points. 


Fig.  387. 


Fig.  38S. 


094.  Two  quadrilaterals  are  equivalent  when 
their  diagonals  are  respectively  equal,  and  form 
equal  angles. 


09o.  If,  on  the  hjrpotenuse  and  sides  of  a  right 
angled  triangle,  semicircumferences  be  described, 
that  upon  the  hypotenuse  passing  through  the  ver- 
tex, the  sum  of  the  crescents  thus  formed  will  be 
equal  to  the  area  of  the  triangle. 


THEOEEMS  IN  SPECIAL  GEOMETRY. 


257 


696.  The  bisectors  of 
any  two  exterior  angles 
of  a  triangle  meet  in  a 
point  which  is  the  centre 
of  a  circle,  to  which  one 
side  of  the  triangle  and 
the  other  two  produced 
are  tangents. 

These  circles  are  called 
the  escribed  circles. 


Fig.  3S9. 


THE  NINE  POINTS  CIRCLE 

697.  In  any  triangle  the  centres  of 
the  THREE  sides,  the  feet  of  the  three 
perpendiculars  from  the  vertices  upon 
the  opposite  sides  or  sides  produced, 
and  the  three  middle  points  of  the 
distances  from  the  vertices  to  the 
common  intersection  of  the  perpen- 
diculars, are  nine  points  in  the  circum- 
ference of  one  and  the  same  circle ;  the 
centre  of  this  circle  is  at  the  middle  of 
the  line  joining  the  centre  of  the  cir- 
cumscribed circle  and  the  common  intersec- 
tion of  the  perpendiculars ;  and  the  radius  is 
half  the  radius  of  the  circumscribed  circle. 

Sug's. — The  student  will  do  well  to  confine  his  at- 
tention in  the  first  instance  to  the  first  figure,  and 
after  he  sees  the  demonstration  in  this  case — i.  e., 
when  the  perpendiculars  fall  within,  to  trace  it  in 
the  case  of  an  obtuse  angled  triangle,  in  which  the 
perpendiculars  fall  on  the  sides  produced 

1st.  To  show  that  the  circle  which  passes  through 
a,  b,  and  c,  also  passes  through  a,  we  show  the  fol- 
lowing relations  among  the  angles :    cab  =  cAb  = 


Fig.  300. 


Fig.  391. 


258  EXERCISES  IN  GEOMETRICAL  XNVENTION. 

cka  +  akh  =■  cah.  Hence,  the  vertices  a  and  a  are  in  the  same  circumference. 
In  like  manner  we  show  that  ft  and  y  are  in  the  circumference  passing  through 
a,  h,  and  c. 

2d.  Considering  one  of  the  partial  triangles  as  BHC,  a,  /?,  and  y  are  the  feet  of 
the  three  perpendiculars  from  its  vertices  upon  one  of  its  sides  and  the  prolonga- 
tion of  the  other  two.  Therefore,  by  the  first  part  r  and  q  are  the  middle  points 
of  CH  and  BH.  Considering  either  of  the  other  partial  triangles  we  find  p  the 
centre  of  AH, 

3d.  ace  and  hft  being  chords  of  the  nine  points  circle,  0  is  its  centre,  and  let- 
ting Q  be  the  centre  of  the  circumscnbed  circle,  we  may  readily  show  that  0 
is  ill  QH,  and  also  is  at  its  middle  point. 

4th.  Drawing  aO,  and  producing  it,  we  maj^  show  that  it  intersects  AH  in  j9, 
and  hence  ^>H  =  A^;  =  Qa,  and  kOiap  is  a  parallelogram.  Therefore  0^;  =  ^pa  = 
\Qik. 

608.  Cor. — The  middle  points  of  the  three  lines  joining  the  cen- 
tres, two  and  two,  of  the  escribed  circles  of  a  triangle,  and  the  middle 
points  of  the  three  lines  joining  the  centres  of  the  escribed  circles 
with  the  centre  of  the  inscribed  circle,  are  six  points  in  the  circum- 
ference of  the  circle  circumscribed  about  the  same  triangle. 


090.  If  one  triangle  be  inscribed  in  another,  the  circumferences 
circumscribing  the  three  exterior  triangles  thus  formed  intersect  in 
a  common  point. 

SuG. — The  demonstration  is  founded  on  the  property  of  the  opposite  angles 
of  an  inscribed  quadrilateral.  The  construction  lines  extend  from  the  vertices 
of  the  inscribed  triangle  to  the  intersection  to  be  examined. 


700.  The  difference  between  the  hypotenuse  and  the  sum  of  the 
other  two  sides  of  a  right  angled  triangle  is  equal  to  the  diameter  of 
the  inscribed  circle. 


701.  If  from  the  extremities  of  any  side  of 
a  triangle  two  lines  be  drawn,  one  bisecting  an 
interior  and  the  other  an  exterior  angle,  these 
lines  will   meet  if  sufficiently  produced,  and 
their  included  angle  Avill  be  half  the  third  angle  of  the  triangle. 

702.  An  inscribed  equilateral  triangle  is  one-fourth  the  circum- 
scribed equilateral  triangle  about  the  same  circle. 


703.  The  three  altitudes  of  a  triangle  are  to  each  other  inversely 

as  the  sides  upon  which  they  fall. 


THEOREMS  IN   SPECIAL  GEOMETRY. 


259 


704.  The  bisectors  of  the  angles 
included  by  the  opposite  sides  of  an 
inscribed  quadrilateral  intersect  at 
right  angles. 

SUG.— By  means  of  {214:)  show  that  FC  + 
CE  +  HA  +  AG  =  180°.  Whence  FOE  =  00^ 


705.  Two  triangles  which  have  an 
angle  in  each  equal,  are  to  each  other 


Fig 


as  the  rectangle  of  the  sides  including  the  equal  angle. 


Sug's.  a  and  D  being  equal,  we  are  to  show- 
that  ABC  :  DEF  : :  AB  x  AC  :  DE  x  DF.  Take  AE' 
=  DE,  AF'  =  DF,  and  draw  E'F'.  Now  from  the 
facts  tliat  the  triangles  AE'F'  and  DEF  are  equal,  and 
that  triangles  of  the  same  altitudes  are  to  each  other 
as  their  bases,  t-he  proposition  is  proved. 


Fig.  394 


706.  If  of  the  four  triangles  into  which  the  diagonals  divide  a 
quadrilateral,  two  opposite  ones  are  equivalent,  the  figure  is  a 
trapezoid. 


707.  The  difference  between  the  angles  which  a  bisector  in  a 
triangle  makes  with  the  side  to  which  it  is  drawn,  is  equal  to  the 
difference  of  the  angles  of  the  triangle  including  this  side. 


708.  If  any  number  of  equal  right  lines  radiate  from  a  common 
point,  making  equal  consecutive  angles,  and 

any  line  be  drawn  through  the  common 
point,  the  sum  of  the  perpendiculars  upon 
this  line  from  the  extremities  of  the  radiat- 
fng  lines  on  one  side,  is  equal  to  the  sum  of 
those  on  the  other  side. 

709.  Cor. — In  any  regular  polygon,  the  ^^^  3^5 
sum  of  the  perpendiculars  let  fall  from  the 

vertices  on  the  one  side  of  any  line  passing  through  the  centre,  is 
equal  to  the  sum  of  those  let  fall  from  the  vertices  on  the  other 
side. 


*' 

'frT\^ 

d 

c' 

\tl'..\ 

^ 

v 

\-^ 

n-/ 

b 

a 

ec:=».,r^ 

a 

260  EXERCISES  IN  GEOMETRICAL  INVENTION. 

7 10.  If  the  sum  of  two  opposite  sides  of  a  quadrilateral  is  equal 
io  the  sum  of  the  other  two  opposite  sides,  a  circle  may  be  inscribed 
in  it. 

Sug's. — Bisect  any  two  adjacent  angles,  as  A  and  B. 

Then  are  the  perpendiculars  r,  r,  r,  equal  (?);    and  it 

remains  to  be  shown  that  the   perpendicular  p  =  r. 

Take  AD'  =  AD,  and  BC  ==  BC,  and  draw  OD'  and  OC. 

Since  a  +  h  =  c  +  d,  and  CD'  =  h  —  {b  —  c)  —  {p  —  d), 

CD'  =  «,  and  the  triangles  DOC  and  D'OC  are  equal. 

Hence  p  =  r. 
Fig.  3%.  ^ 


711.  If  two  planes  are  parallel,  any  right  line  which  pierces  one. 


pierces  the  other  also. 
SuG. — Proof  based  on  {410). 


712.  If  two  planes  are  parallel,  any  plane  which  intersects  one, 
intersects  the  other  also,  and  the  lines  of  intersection  are  parallel. 

713.  Cor. — Two  planes  which  are  parallel  to  a  third,  are  parallel 
to  each  other. 


7 14.  A  plane  which  is  perpendicular  to  a  line  of  another  plane, 
or  to  a  line  parallel  to  that  plane,  is  itself  perpendicular  to  the 
latter  plane. 


7 IS.  If  a  straight  line  is  perpendicular  to  a  plane,  any  line  par- 
allel to  the  plane  is  perpendicular  to  the  first  line. 

SuG. — Two  lines  in  space  which  are  not  in  the  same  plane,  are  said  to  make 
the  same  angle  with  each  other  as  two  lines  respectively  parallel  to  them  and 
both  lying  in  one  plane. 


716.  In  order  that  a  straight  line  be  perpendicular  to  a  plane,  it 
is  sufficient  that  it  be  perpendicular  to  two  lines  not  parallel  to  each 
other,  and  situated  in  the  plane  or  parallel  to  it. 


717.  If  two  right  lines  in  space  are  perpendicular  to  each  other 
(not  necessarily  intersecting),  their  projections  on  a  plane  parallel 
to  either  line  are  perpendicular  to  each  other. 

SxjG.— The  Pi'qjeciion  here  referred  to  is  that  which  is  called  the  OriJio- 
graphic  Projection.  The  proposition  is  not  generally  true  of  the  Perspective 
Projection,  i.  e.,  the  spaces  which  the  lines  (considered  as  material)  would  appear 
to  occupy  if  they  were  placed  between  the  eye  and  the  plane.  (See  Ex.  8, 
page  174,  Part  II.) 


THEOREMS   IN  SPECIAL  GEOMETRY. 


261 


718.  The  angle  of  inclination  {392,  Part  II.)  of  a  line  oblique 
to  a  plane,  is  less  than  the  angle  included  be- 
tween   this   line   and   any   line  of   the  plane, 
except  its  projection,  which  passes  through  the 
point' in  which  the  first  line  pierces  the  plane. 


SuG.  BD  being  the  projection 
ABD',  BD'  being  any  line  other 
thi"ough  B. 


719'  Between  any  two  lines  not  in  the  same  plane,  one  line,  and 
only  one,  can  be  drawn,  which  shall  be  perpendicular  to  both  the 
given  lines. 

Sug's. — Pass  a  plane  through  one  of  the  lines  parallel  to  the  other;  and 
through  the  other  line  pass  a  plane  perpendicular  to  the  first  plane. 


of   AB,    ABD    <     / 
than  BD,   passing 

Fig.  3t)T. 

720.  In  a  warped  quadrilateral,  i.  e.,  one  whose  sides  do  not  all 
lie  in  the  same  plane,  the  middle  points  of  the  sides  are  in  one 
plane,  and  are  the  vertices  of  the  angles  of  a  parallelogram. 

SuG. — Conceive  the  planes  of  two  opposite  angles  of  the  quadrilateral,  the 
intersection  of  which  will  be  a  diagonal  of  the  given  quadrilateral. 


721.  A  line  being  given  in  a  plane,  one  plane  can  be  drawn  in- 
cluding the  given  line  and  perpendicular  to  the  first  plane,  and  only 
one.     Hence  all  right  diedrals  are  equal. 

SuG. — Demonstration  similar  to  {390). 


722.  The  plane  angle  formed  by  drawing  two  lines  in  the  faces 
of  a  diedral,  from  a  common  point  in  the  edge,  is  less  than  tlie 
measure  of  the  diedral 
if  the  angle  is  acute, 
and  lines  lie  on  the 
same  side  of  the  plane 
of  the  measure  and 
are  equally  inclined  to 
it,  and  greater  if  they 
lie  on  opposite  sides. 

Li  general,  if  the 
diedral  is  acute,  the 
limits  of  the  varying 
angle  are  0  and  90° 
when  both  the  Imes  lie 


^-^^ 

o 

A 

p>^ 

>^^^^ 

K^D-::::; 

>^    r 

ker 

'6' 

■^^*^A 

(«) 


■■••.. 

^a' 

\ 


Fig.  398. 


262  EXERCISES    I2f    GEOMETRICAL    I>rVE2fTI0:N". 

on  the  same  side  of  the  plane  of  the  measuring  angle,  and  180°  when 
they  lie  on  opposite  sides.  If  the  diedral  is  obtuse  the  limits  are  0 
and  the  angle  itself  when  the  lines  lie  on  the  same  side  of  the 
measure,  and  90°  and  180°  when  they  lie  on  opposite  sides. 


0^\., 


'23.  If  the  projections  of  a  line  in  the  two  faces  of  a  diedral  are 
straight,  the  line  is  a  straight  line. 

SuG.— Proof  based  on  {386). 


724.  If  from  the  vertex  of  a  triedral  a  line  be  drawn  at  pleasure 
within  the  triedral,  the  sum  of  the  plane  angles  formed  by  this  line 
and  any  two  edges  is  less  than  the  sum  of  the  facial  angles  formed 
by  the  other  edge  and  these  two. 


72o.  If  through  a  point  in  space  two  lines  be  drawn  parallel  to  a 
given  plane,  and  through  the  same  point  two  planes  be  passed  re- 
spectively peq^endicular  to  the  two  lines,  the  intersection  of  these 
two  planes  will  be  perpendicular  to  the  given  plane. 


-^   726.  The  three  planes  which  bisect  the  three  diedrals  of  a  trie- 
dral intersect  in  a  common  line. 


727.  In  any  convex  polyedral,  the  sum  of  the  diedrals  is  greater 
than  the  sum  of  the  angles  of  a  polygon  having  the  same  number 
of  sides  that  the  polyedral  has  faces. 

Srr;.— Proof  based  upon  {T22). 


728.  Def. — A  JPoIyedroii  is  a  solid  bounded  by  plane  sur- 
faces. A  RegyJar  Convex  Pohjcdron  is  a  polyedron  whose  faces  are 
all  equal  regular  polygons,  and  each  of  whose  solid  angles  is  con- 
vex outward,  and  is  enclosed  by  the  same  number  of  faces. 


729.  There  are  five  and  only  five  regular  convex  polyedrons — viz. : 
The  Tetraedron,  whose  faces  are  four  equal  equilateral  triangles  ;  The 
Hexaedron,  or  Cule,  whose  faces  are  six  equal  squares;  The  Ocfaedro?i, 
whose  faces  are  eight  equal  equilateral  triangles  ;  The  DodecaedroUf 
whose  faces  are  twelve  equal  regular  pentagons ;  and  Tlie  Icosaedron, 
whose  faces  are  twenty  equal  equilateral  triangles. 


THEOREMS   IN  SPECIAL  GEOMETRY. 


263 


Dem. — We  demonstrate  this  proposition  by  showing — 1st,  that  such  solids 
can  be  constructed  ;  and  2d,  that  no  others  are  possible. 

The  Regular  Tetraedron. — Taking  three  equal  equilateral  ti'iangles,  as  ASB,  ASC 
and  BSC,  it  is  possible  to  enclose  a  solid  angle,  as  S,  with 
them,  since  the  sum  of  the  three  facial  angles  is  (what  ?) 
(Part  II.,  480).  Then,  since  AC  =  AB  =  CB  (?),  consid- 
ering ACB  the  fourth  face,  we  have  a  regular  polyedroQ 
whose  four  faces  are  equilateral  triangles. 

The  Reguhr  Hexaedron  or  Cuhe. — This  is  a  familiar 
solid,  but  for  purposes  of  uniformity  and  completeness  we 
may  conceive  it  constructed  as  follows  :  Taking  three 
equal  squares,  as  ASCB,  CSED,  and  ASEF,  we  can  en- 
close a  solid  angle,  as  S,  with  them  (?).  Now,  conceive 
the  planes  of  CB  and  CD,  AB  and  AF,  EF  and  ED  pro- 
duced. The  plane  of  CB  and  CD  being  parallel  to  ASEF  (?) 
will  intersect  the  plane  of  EF  and  ED  in  HD  parallel  to 
FE  (?).  In  like  manner  FH  can  be  shown  parallel  to  ED, 
BH  to  CD,  and  HD  to  BC.  Hence  the  solid  has  for  its 
faces  six  equal  squares. 


Fig.  399. 


A 

/ 

! 

/H 

7 

c 

Fig.  400. 


Fig.  401. 


Those  triedrals  are 


The  Regular  Octaedron. — At  the  intersection  P,  of  the 
diagonals  of  a  square,  ABCD,  erect  a  perpendicular  SP  to 
the  plane  of  the  square,  and  making  SP  =  AP  (half  of  one 
of  the  diagonals)  draw  SA,  SD,  SC,  and  SB.  Making  a 
similar  construction  on  the  olhei"  side  of  the  plane  ABCD, 
we  have  a  solid  having  for  faces  eight  equal  equilateral 
triangles. 


The  Regular  Dodecaedron. — Taking  twelve  equal  regu- 
lar pentagons,  it  is  evident  that  we  may  group  them  in 
two  sets  of  six  each,  as  in  the  figure.  Thus,  around  0  we 
may  place  five,  forming  5  triedrals  at  the  vertices  of  O. 
possible,  since  the  sum  of  the  facial  angles  enclosing  each  is  3|  right  angles  (?) 
—i.  e., between  0  and  4  right  angles  (Part  II.,  436).  In  like  manner  the  other 
6  may  be  grouped  by  placing  5  of  them  about  0'.  Now,  conceiving  the  convexity 
of  the  group  O  in  front  and  the  con- 
cavity of  group  O',  we  may  place 
tl»e  two  together  so  as  to  inclose  a 
solid.  Thus,  placing  A  at  h,  the 
three  faces  5,  6,  1,  will  enclose  a  tri- 
edrul,  since  the  diedral  included  by 
5  and  1  is  the  diedral  of  such  a  tri- 
edral.  Then  wnll  vertex  B  fliU  at 
c,  and  a  like  triedral  will  be  formed 
at  that  point,  and  so  of  all  the  other 

vertices.     Hence  we  have  a  polyedron  having  for  faces  12  equal  regular  penta- 
gons. 


Fig.  402. 


264 


EXERCISES  IN  GEOMETRICAL  INVENTION. 


Fig.  403. 


Tlie  Regular  Icosaedron. — Taking 
20  equal  equilateral  triangles,  they 
can  be  grouped  in  two  sets,  as  in 
tlie  figure,  in  a  manner  alt6getber 
similar  to  the  preceding  case.  The 
solid  angles  in  this  case  are  included 
by  5  facial  angles  whose  sum  is  3^ 
right  angles  (?),  which  is  a  possible 
case  (Part  II.,  436).  *As  before, 
conceiving  the  convexity  of  group 
O  in  front,  and  the  concavity  of  O',  we  can  place  them  together  by  placing  A  at 
a,  tlms  enclosing  a  solid  angle  with  5  faces,  whence  B  will  fall  at  6,  etc.  Thus 
>ve  obtain  a  solid  with  20  equal  equilateral  U'iangles  for  its  faces. 

That  there  can  be  no  other  regular  polyedrons  than  these  5  is  evident,  since  we 
can  form  no  other  convex  solid  angles  by  means  of  regular  polygons.  Thus,  with 
equilateral  triangles  (the  simplest  polygon)  we  have  formed  solid  angles  witli  3 
faces  (the  least  number  possible),  as  in  the  tetraedron ;  with  4,  as  in  the  octaedron  ; 
and  with  5,  as  in  the  icosaedron.  Six  such  facial  angles  cannot  enclose  a  solid 
angle,  since  their  sum  is  four  right  angles  (?),  and  much  less  any  gi-eater  num- 
ber. Again,  with  squares  (the  next  most  simple  polygon)  we  have  formed  solid 
angles  with  3  faces  as  in  the  hexaedron,  and  can  form  no  other,  for  the  same 
reason  as  above.  With  regular  pentagons  we  can  only  enclose  a  triedral,  as  in 
the  dodecaedron,  for  a  like  reason.  With  regular  hexagons  we  cannot  enclose 
a  solid  angle  (?),  and  much  less  with  any  regular  polygon  of  more  than  six 
sides. 

ScH.— Models  of  the  regular  polyedrons  are  easily  foi-med  by  cutting  the  fol- 
lowing figures  from  cardboard,  cutting  half-way  through  the  board  in  the  dotted 
lines,  and  bringing  the  edges  together  as  the  forms  will  readily  suggest. 


Fifi.  404. 


THEOEEMS   IN  SPECIAL  GEOMETRY. 


265 


730.  Any  regular  polyedron  is  iuscriptible  and  circumscriptible 
by  a  sphere. 

Sug's.— From  the  centres  of  any  two  adjacent  faces,  as  c  and  c',  let  fall  per- 
jjendiculars  upon  the  common  edge,  and  they  will  meet  it  in 
tlie  same  point  o  (V).  The  plane  of  these  lines  will  be  per- 
pendicular to  this  edge  (?),  and  perpendiculars  to  these  faces 
from  their  centres,  as  cS,  c'S,  will  lie  in  this  plane  (?),  and  hence 
will  intersect  at  a  point  equally  distant  from  these  faces. 

In  hke  manner  c"S  =  c'S,  and  the  point  S  can  be  shown  to 
be  equally  distant  from  all  of  the  faces,  and  is  therefore  the 
centre  of  the  inscribed  sphere. 

Joinmg  S  with  the  vertices,  we  can  readily  show  that  S  is 
also  the  centre  of  the  circumscribed  sphere. 


Fiu.  405. 


731.  Show  that  a,  being  the  edge  of  a  regular  tetraedron,  its 


volume  IS  — ^— 


732.  r>EF. — A  Truncated  Prism  is  one  whose  upper  and 
loAver  bases  are  not  parallel. 

733.  The  volume  of  a  truncated  triangular  prism  is  equal  to  the 
sum  of  the   volumes  of  three  pyramids  -' 
whose  common  base  is  the  lower  base  of 
the  prism,  and  Avhose   vertices  are  the 
angles  of  the  upper  base. 

Sug's. — Let  bD,  cD',  and  «D"  be  perpendicu- 
lar to  the  lower  base.  Volume  of  6-ABC  is  ^  6D 
X  ABC.  Volume  «-6Cc  :  volume  6-ABC  ::  cb'C 
:  bBC  :  :  cC  :  b's  :  :  cD'  :  bb.  .'.  Volume  a-bC(}' 
—  ^cD'  X  ABC.  In  a  similar  manner  volume 
b-a'AC  =  iaD"  x  ABC. 

Fig.  406. 

^  734.  Cor. — The  volume  of  a  prism, 

one  of  whose  bases  is  a  right  section  and  the  other  an  oblique 
section,  is  the  product  of  the  right  section  into  the  arithmetical 
mean  of  its  edges. 

Sug's.— The  volume  of  a&c-ABC  is  as  shown  above  ABC  I j. 

But  if  ABC  is  a  right  section,  bD  =  bB,  cD'  =  cC,  and  aD"  =  aA.  Hence  the 
volume  is  ABC  ^^B  +  cC   +  gAJ 


266  EXERCISES    IN  GEOMETRICAL  INVENTtON. 

73o.  The  volume  of  any  polyedi'on  having  for  its  bases  any 
two  polygons  whatever,  situated  in  parallel  planes,  and  for  lateral 
faces  trapezoids,  is  the  product  of  -J-  the  distance  between  the  bases 
into  the  sum  of  the  two  bases  plus  -i  times  a  section  midway  be- 
tween the  bases;  or  v  =  ^  (b  +  B'  +  4b"),  in  which  H  is  the  dis- 
tance between  the  bases,  B  and  B'  the  bases,  and  b"  a  section  mid- 
way between  the  bases. 

l'     ,'      ^  Dem. — Let  Li  Ml  Ni  Pi  Qi  be  the  section  of  such  a 

p,      polyedron  midway  bet-ween  its  bases,  and  S  any  point 

\        in  this  section.    Joining  S   with   the  vertices  of  the 
^-^-Ni    polyedron,  we  divide  the  solid  into  as  many  pyramids 
as  it  has  faces.     The  volumes  of  the  two  which  have 
B  and  B'  for  their  bases  are  evidently  ^H  x  B,  and  ^H 
X  B'.     It  remains  to  find  the  volume  of  the  others. 
J,     ^-  Let  LML'M'  be  a  lateral  face  corresponding  to  LiMi 

and  SO  a  perpendicular  from  S  upon  this  face.  Draw 
ri  through  0  perpendicular  to  LM,  and  consequently  to  L'M'.  Take  I'Ki  per- 
pendicular to-the  plane  section,  whence  I'Ki  —  |H.  Kow  the  volume  of  the 
pyramid  having  L'M'LM  for  its  base  and  S  for  its  vertex  is  LiMi  x  2rii  x  ^SO. 
But  I'll  X  SO  =  Sli  X  I'Ki  (?) ;  whence  the  volume  of  this  pyramid  is  f  LiMi  x 
SI.  X  I'Ki  =  I  X  2SLiM,  x  I'Ki  =  ^  I'K,  x  4SLiMi  =  ^H  x  4SLiMi.  In  like 
manner  the  volume  of  the  pyramid  having  for  its  base  the  face  in  which  M^  N^ 
is  situated,  can  be  shown  to  be  ^H  x  4SMi  Ni  and  similarly  of  all  the 
others.    Whence  the  whole  volume  is  ^  H  (B  +  B'  +  4B"). 

736.  Cor. — The  proposition  is  equally  true  when  some  or  all  of 
the  lateral  faces  are  triangles ;  i.  e.,  wheu  one  base  has  more  sides 
than  the  other. 

ScH. — The  preceding  propositions  are  of  much  value  in  calculating  earth- 
work. 


737.  If  we  cut  a  pyramid  by  a  plane  parallel  to  its  base,  a  second 
pyramid  is  formed  similar  to  the  first. 


738.  Two  triangular  pyramids  are  similar  wiienever  they  have 
an  equal  diedral  angle  contained  between  faces,  similar  each  to  each, 
and  similarly  placed. 


739.  Two  polyedrons  composed  of  the  same  number  of  tetrae- 
drons,  similar  each  to  each,  and  similarly  disposed,  are  similar. 


PROBLEMS  IN   SPECIAL  OR  ELEMENTARY   GEOMETRY.  267 

740.  All  regular  polyedrous  of  the  same  number  of  faces  are 
similar  solids. 


74:1.  The  intersection  of  the  surfaces  of  two  spheres  is  the  cir- 
cumfer'euce  of  a  circle  whose  plane  is  perpendicular  to  the  line  which 
joins  their  centres. 


742.  Through  any  four  points  not  in  the  same  plane  one  sphere 
may  be  made  to  pass,  and  only  one. 

Sug's.— The  four  points  may  be  considered  as  the  vertices  of  a  tetraedron. 
Conceive  perpendiculars  drawn  to  the  triangular  faces  from  the  intersections 
of  lines  drawn  in  these  faces  perpendicular  to  the  sides  at  their  middle  points. 
These  perpendiculars  will  meet  at  a  common  point  (?),  which  is  the  centre  of 
the  circumscribed  sphere  (?). 

[The  student  should  show  why  only  one  sphere  can  be  circumscribed.] 

743.  OoR.  1. — The  four  perpendiculars  erected  at  the  centres  of 
the  circles  circumscribing  the  faces  of  a  tetraedron  intersect  at  a 
common  point. 

744.  Cor.  2. — The  six  planes,  perpendicular  to  the  six  edges  of 
a  tetraedron  at  their  middle  points,  intersect  at  the  centre  of  the 
circumscribed  sphere. 


745.  One  sphere  and  only  one  may  be  inscribed  in  any  tetraedron. 
SuG. — Bisect  the  diedrals  with  planes. 


746.  The  angle  included  by  any  two  curves  intersecting  on  the 
surface  of  a  sphere,  is  equal  to  the  angle  included  by  the  arcs  of  two 
great  circles  passing  through  tlif  point  of  intersection,  and  whose 
planes  produced  include  the  tangents  to  the  curves  at  their  inter- 
section. 


SECTION  IL 

PROBLEMS  IN  SPECIAL  OR  ELEMENTARY  GEOMETRY. 


747.  To  bisect  the  angle  formed 
l)y  two  lines  whose  intersection  is 
inaccessible. 

Suo. — M  and  N  are  points  in  the  bisector. 


Fig.  40S. 


268 


EXERCISES  IN  GEOMETRICAL   INVENTION. 


748.  To  pass  a  circumfereDce  through  three  points,  not  in  the 
same  straight  line,  when  the  radius  is  so  long  as  to  render  the  ordi- 
nary method  impracticable. 

SuG. — Let  A,  B,  and  C  be  the  three  points ;  then  are  M  and  N  other  points  in 
the  same  circumference. 


Fig.  409. 

■N       749.  From  two  given  points  on  the  same 
side  of  a  line  given  in  position,  to  draw  two 
lines  which  shall  meet  in  that  line  and  make 
Ej   /tfK/g  equal  angles  with  it. 

i     A 

i    /  SuG. — K  a  and/?  are  equal,  what  is  the  relation  of 

!/  MEtoEF? 

Pig.  410. 


750.  To  construct  an  isosceles  triangle  with  a  given  base  and 
vertical  angle. 

Sua.— See  Prob.  4,  p.  103. 


7S1.  To  trisect  a  right  angle. 

Sua.— What  is  the  value  of  an  angle  of  an  equilateral  triangle  ? 


7S2.  Given  the  perpendicular  of  an  equilateral  triangle,  to  con- 
struct the  triangle. 


7o3.  Given  the  diagonal  of  a  square,  to  construct  it. 


7o4.  To  construct  an  isosceles  triangle,  so  that  the  base  shall  be 
a  given  line,  and  the  vertical  angle  a  right  angle. 


PROBLEMS   IN   SPECIAL  OR  ELEMENTARY   GEOMETRY. 


269 


755'  Given  the  sum  of  the  diagonal  and  a  side 
of  the  equare,  to  construct  it. 

SuG.— What  are  the  values  of  a  and  (5  respectively? 


756.  To  construct  a  triangle  when  the  altitude, 
the  vertical  angle,  and  one  of  the  sides  are  given. 


Fig.  411. 


757.  To  construct  a  triangle  when  the  sum  of  the  three  sides 
and  the  angles  at  the  base  are  given. 


Sug's.— MN  being  the  sum  oia.h,  and 
c,  what  are  the  angles  M  and  N  as  com- 
pared with  the  given  angles  a  and  fi  ? 


Fig.  412. 


758.  In  a  right  angled  triangle  the  perimeter,  and  the  perpen- 
dicular from  the  right  angle  upon  the  hypotenuse  being  given,  to 
construct  the  triangle. 


Bug's.— DE  is  equal  to  the  perimeter, 
DBE  is  an  angle  of  135°,  and  FE  is  the 
perpendicular  on  the  hypotenuse.  ABC 
is  the  required  triangle.  Let  the  student 
give  the  solution  in  full,  and  the  proof 


759.  From  two  given  points  on  the  same  side 
of  a  given  line,  to  draw  two  equal  straight  lines 
which  shall  meet  in  the  same  point  of  the  line. 


700.    To  pass  a  circumference  through  two 
gfv'en  points,  which  shall  have  its  centre  in  a  given  line. 


76*1.  To  construct  a  quadrilateral  when  three  sides,  one  angle, 
and  the  sum  of  two  other  angles  are  given. 

Sug's.— What  is  the  fourth  angle  ?  When  two  sides  and  their  included  angle 
are^nown,  there  will  be  two  cases,  according  as  the  two  angles  whose  sum  ia 
known  are  adjacent  to  each  other  or  opposite.  In  the  latter  case  we  have  to 
describe  a  segment  on  a  diagonal,  which  will  contain  the  fourth  angle.  For  the 
third  case  see  Ex.  13,  page  136. 


270  EXZEasES  rs  geometpjcal  inyen-tion. 


762.  To  construct  a  quadrilateral  when  three 
angles  and  two  opposite  sides  are  given. 


763.    To   bisect  a   trapezoid    by   a    line 
drawn  from  one  of  its  angles. 


Fig.  416. 


764.  In  a  given  circle,  to  inscribe  a  triangle  equiangular  with  a 
given  triangle. 

SuG. — How  does  an  angle  at  the  centre   compare  with  one  inscribed  in 
the  same  segment  ? 


76o.  To  describe  three  circles*  of  equal  diameters  which  shall 
touch  each  other. 


766.  In  an  equilateral  triangle,  to  inscribe  three  equal  circles 
which  shaU  touch  each  other  and  the  three  sides  of  the  triangle. 

767.  To  describe  a  circle  of  given  radius  touching  the  two  sides 
of  a  given  angle. 

SuG.— How  far  is  the  centre  from  each  line  ? 


768.  To  describe  a  circumference  which  shall  be  embraced  be- 
tween two  parallels  and  pass  through  a  given  point  within  the  par- 
allels. 

SuG.— In  what  hne  Is  the  centre  ?    How  far  from  the  given  point  ? 


769.    To  describe  a  circle  with  a  given  radius,  which  shall  pass 
through  a  given  point  and  be  tangent  to  a  given  line. 


770.  To  find  in  one  side  of  a  triangle  the  centre  of  a  circle  which 
shall  touch  the  other  two  sides. 


771.  Through  a  given  ]X)int  on  a  circumference,  and  another 


PROBLEMS   IN    SPECIAL  OR  ELEMENTARY   GEOMETRY.  271 

given  point  without,  to  describe  a  circle  touching  the  given  circum- 
ference. 

SuG. — Consider  in  what  two  lines  the  centre  must  h'e. 


772.  In  the  diameter  of  a  circle  produced,  to  determine  the  point 
from  which  a  tangent  drawn  to  the  circumference  shall  be  equal  to 
the  diameter. 

SuG. — What  is  the  relation  between  the  radius,  the  required  tangent,  and  the 
distance  from  tlie  centre  to  the  intersection  of  the  produced  diameter  and  the 
required  tangent? 


775.  To  describe  a  circle  of  given  radius;  touching  two  given 
circles. 


774.  In  a  given  circle,  to  inscribe  a  right  angle,  one  side  of  which 

is  eriven.  -  ,     •\ 

.    /;?] 

77 5.  In  a  given  circle,  to  construct  an  inscribed  triangle  of  given 
altitude  and  vertical  angle. 


776'  To  inscribe  a  square  in  a  given  right- 
angled  isosceles  triangle,  one  side  being  in  tlie 
hypotenuse. 


Fig.  4r 


777.  To  inscribe  a  square  in  a  given  quadrant  of  a  circle,  the 
vertex  of  an  angle  being  at  the  centre. 

778.  To  find  the  centre  of  a  circle  in  which  two  given  lines 
meeting  in  a  point  shall  be  a  tangent  and  a  chord. 


779.  To  describe  a  circumference  which  shall  pass  through  a 
given  point  and  be  taugent  to  a  given  line  at  a  given  point. 


780.  To  bisect    a  quadrilateral  by  a  line 
drawn  from  one  of  its  angles. 

SuG. — The  demonstration  is  based  upon  the  prin- 
ciple that  triangles  having  equal  bases  and  equal      [^ 
altitudes  are  equivalent.  *'^     U   pj^  ^^q 


272  EXERCISES   IN   GEOMETRICAL  INVENTION. 

781.  Through  a  given  point  situated  between  the  sides  of  an 
angle,  to  draw  a  line  terminating  at  the  sides  of  the  angle,  and  in 
such  a  manner  as  to  be  bisected  at  the  point, 

SuG. — Conceive  the  point  as  situated  in  the  third  side  of  a  triangle  of  which 
the  two  given  lines  are  the  other  two. 


782.  To  draw  a  line  parallel  to  the  base  of 
a  triangle  so  as  to  divide  the  triangle  into  two 
equivalent  parts. 

SuG.     PB'  =  DB'  =  iAB'.    See  {344,362). 


Fig.  419. 


783.  To  construct  a  square  when  the  difference 
between  the  diagonal  and  a- side  is  given. 

SuG. — Consider  the  angles. 


Fig.  420. 

784.  To  determine  the  point  in  the  circumference  of  a  circle 
from  which  chords  drawn  to  two  given  points  shall  have  a  given 
ratio. 

SuG. — Draw  a  chord  dividing  the  chord  joining  the  given  points  in  the  re- 
quired ratio,  and  bisecting  one  of  the  subtended  arcs. 


78S.  To  bisect  a  given  triangle  by  a  line  drawn  from  one  of  its 
angles. 

786.  To  bisect  a  given  triangle   by  a  line  drawn 
yi/A     from  a  given  point  in  one  of  its  sides. 


Fig.  421 


787.  In  the  base  of  a  triangle  find  the  point  from  which  lines 
extending  to  the  sides,  and  parallel  to  them,  will  be  equal. 


788.  To  construct  a  parallelogram  having  the  diagonals  and  one 
side  given. 

789.  To    construct  a  triangle    when  the    three  altitudes   are 
given. 


PROBLEMS   IN   SPECIAL   OR   ELEMENTARY   GEOMETRY.  273 

SxjG.— What  is  the  relation  of  the  perpendiculars  to  the  sides  upon  which 
they  fall?  If  a  triangle  can  be  formed  with  the  perpendiculars  as  sides,  how 
will  it  compare  with  the  first  triangle  ?  How  proceed  when  the  perpendiculars 
will  not  form  a  triangle  ? 


790'  What  is  the  area  of  the  sector  whose  arc  is  50°,  and  whose 
radius  is  10  inches? 


701-  To  construct  a  square  equivalent  to  the  sum,  or  to  the  dif- 
ference of  two  given  squares. 

702.  To  divide  a  given  straight  line  in  the  ratio  of  the  areas  of 
two  given  squares. 

703.  To  construct  a  triangle,  when  the  altitude,  the  line  bisecting 
the  vertical  angle,  and  the  line  from  the  vertex  to  the  middle  of  the 
base  are  given. 

SuG. — The  centre  of  the  circle  circumscribing  the  required  triangle  is  in  the 
perpendicular  to  the  base  at  its  middle  point ;  and  the  intersection  of  this  per- 
pendicular and  the  bisectrix  is  a  point  in  this  circumference. 

Show  that  the  bisector  always  lies  between  the  perpendicular  and  the  medial 
line. 


704:.  Through  a  given  point,  draw  a  line  such  that  the  parts  of 
it,  between  the  given  point  and  perpendiculars  let  fall  on  it  from  two 
other  given  points,  shall  be  equal. 

"What  would  be  the  result,  if  the  first  point  were  in  the  straiglit 
line  joining  the  other  two  ? 


705.  From  a  point  without  two  given  lines,  to  draw  a  line  such 
that  the  part  intercepted  between  the  given  lines  shall  be  equal  to 
the  part  between  the  given  point  and  the  nearest  line. 

SuG. — Produce  the  lines  till  they  meet,  if  necessary.  Draw  a  line  through  the 
given  point  parallel  to  one  of  the  lines,  and  produce  it  till  it  meets  the  other. 


706.  Given  one  angle,  a  side  adjacent  to  it,  and  the  difference  of 
the  other  two  sides,  to  construct  the  trians^le.  c 


Queries.— How  if  Z>  >  a?    How  if  B  is  obtuse? 


v5 


707'  To  pass  a  circumference  through  two  given  p^^  422 

points,  having  its  centre  in  a  given  line. 

18 


274 


EXERCISES   IN    GEOMETRICAL   INVENTION. 


798.  To  draw  a  line  parallel  to  a  given  line  and  tangent  to  a 
given  circumference. 
SuG. — Draw  a  diameter  perpendicular  to  the  given  line. 


799.  To  draw  a  common 
tangent  to  two  given  circles. 

SuG.— 1st  Method.— There  are 
two  sets  of  tangents,  AC,  BD,  and 
A'D',  B'C.  For  the  first,  observe 
that  if  PE  =  AP  -  OC,  OE  is  parallel 
to  AC,  etc. 


Fig.  4-23. 


Fig.  424. 


2d    Method.     pO,    p'O' 
being  parallel  to  each  other, 
pp'T  gives  the  intersection 
of  the  tangent  with  the  line 
passing  through  the  centres, 
since 
OT   :  OT,  or  pO  —p'O'  :  p'O'  :  :  00'  :  O'T. 
Also,  PO  -  P'O'  :  P'O'  :  :  00'^  OT. 
Hence  O'T  is  constant  for  all  positions  of  the  parallel  radii.    Prove  that  if 
the  parallel  radii  are  on  different  sides  of  the  line  joining  the  centres,  T'  is 
the  point  where  the  internal  tangent  cuts  00'. 
Queries. — How  many  tangents  can  be  drawn — 1st.  "When  the  ch'cles  are  ex- 


pO  :  p'O' 


ternal  one  to  the  other; 
they  intersect  each  other  : 
within  the  other  ? 


2d.  When  they  are  tangent  externally;  3d.  When 
4th.  When  tangent  internally;  5th.  When  one  lies 


800.  To  describe  a  circle  tan- 
gent to  a  given  circumference 
and  also  to  a  given  line  at  a  given 
point. 

Sug's. — There  may  be  two  cases — 
1st.  When  the  given  circle  is  exterior 
to  the  one  sought ;  and  2d.  When  it 
is  interior.  In  either  case  the  centre 
of  the  required  circle  is  in  the  perpen- 
dicular AO'.  In  the  former  case,  0,  the 
centre  of  the  required  circle,  is  at 
Fig.  425.  r  +  r'  from  C  ;  and  in  the  latter  0'  is 

at  r  —  r'  from  C.     AO  =  r,  and  CD  =  AB  =  AB'  =  r'. 


PROBLEMS  IN  SPECIAL  OR  ELEMENTARY  GEOMETRY.  275 


801'  To  construct  a  trai^ezoid  when  the 
four  sides  are  given. 

SuG. — Knowing  the  difference  between  the  two 
parallel  sides,  we  may  construct  the  triangle  AEC, 
and  hence  the  trapezoid. 


Fig.  426. 


802.  On  a  given  line,  to  construct  a  polygon  similar  to  a  given 
polygon. 


Sug's. — One  method  may  be  learned  from 
(.90).  Ex.  8,  page  152,  furnishes  another 
method.  The  following  is  an  elegant  method  : 
To  construct  on  A'  homologous  with  A,  a 
polygon  similar  to  P.  Place  A'  parallel  to  A, 
and  the  figure  will  suggest  the  construction. 


Fig.  427 


803.  To  pass  a  plane  through  a  given  line  and  tangent  to  a  given 
sphere. 

Sug's. — Pass  a  plane  through  the  centre  of  the  sphere  and  perpendicular  to 
the  given  line.  Through  the  point  of  intersection  and  in  this  secant  plane  draw 
tangents  to  the  great  circle  in  which  the  secant  plane  intersects  the  surface  of 
the  sphere.  The  points  of  tangency  will  be  the  points  of  tangency  of  the  re- 
quired planes  (?),  of  which  there  are  thus  seen  to  be  two. 


804.  Def. — A  Tanr/ent  I^lane  to  a  cylindrical  or  conical 
surface  is  a  plane  which  contains  an  element  of  the  surface,  but 
does  not  cut  the  surface.  The  element  which  is  common  to  the 
surface  and  the  plane  is  called  the  Element  of  Contact. 


80S.  To  pass  a  plane  through  a  given  point  and  tangent  to  a 
given  cylinder  of  revolution. 

Stjg's. — 1st.  When  the  point  is  in  the  surface  of  the  cylinder.  Through  the 
point  draw  an  element  of  the  cylinder,  by  passing  a  line  parallel  to  the  axis, 
or  to  any  given  element.  Through  the  same  point  pass  a  plane  perpendicular 
trTthis  element,  making  a  right  section  (a  circle).  To  this  circle  draw  a  tan- 
gent. The  plane  of  the  element  and  tangent  is  the  tangent  plane  required. 
[The  student  should  prove  that  any  point  in  the  plane  affirmed  to  be  tangent, 
not  in  the  element  passing  through  the  given  point,  is  without  the  cylinder.] 

2d.  When  the  given  point  is  without  the  cylinder.  Pass  a  plane  through 
the  given  point  perpendicular  to  the  axis  of  the  cylinder,  thus  making  a  right 
section  of  the  cylinder  (a  circle).  In  this  secant  plane  draw  tangents  to  the 
section.  Through  the  points  of  contact  of  these  tangents  draw  elements  of 
the  cylinder.  These  elements  are  the  elements  of  contact  of  the  tangent 
planes:    Hence  planes  passing  through  them  and  the  given  point  are  the  tan- 


276  EXERCISES   IN   GEOMETPJCAL  INVENTION. 

gent  planes  required.    [The  student  should  remember  that  this  is  but  an  ouU 
line,  and  be  careful  to  fill  it  up,  giving  the  proof.] 


806.  To  pass  a  plane  through  a  given  point  and  tangent  to  a 
conical  surface  of  revolution. 


807.  To  find,  with  the  compasses  and  ruler,  the  radius  of  a  ma- 
terial sphere  whose  centre  is  inaccessible. 

Sug's. — With  one  point  of  the  compasses  at  anj'-  point 
in  the  surface,  as  A,  trace  a  circle  of  the  sphere,  as  a.cb. 
The  chord  ka  is  measured  by  the  distance  between  the 
compass  points.  In  like  manner  measure  three  other 
chords,  as  ac,  ab,  and  he.  Draw  a  plane  triangle  having 
these  chords  for  its  sides,  and  circumscribe  a  circle  about 
it.  Thus  «D  is  found.  Knowing  <zA,  and  aD,  and  remem- 
bering that  ArtB  is  right  angled  at  a,  the  triangle 
ArtB  can  be  drawn  in  a  plane  (?},  whence  AO  becomes 
known. 


SECTION   III 

APPLICATIONS  OF  ALGEBRA  TO  GEOMETRY. 

808.  The  mathematical  method  which  is  called  technically  Ap- 
plica fio7is  of  Algebra  to  Geometry  consists  in  finding,  by  means  of 
equations,  the  numerical  values  of  the  unknown  parts  of  a  geomet- 
rical figure,  when  a  sufficient  number  of  the  parts  are  given  numeri- 
cally. 

809.  By  reference  to  the  Complete  School  Algebra,  page 
238,  it  will  be  seen  that  the  algebraic  solution  of  a  problem  consists 
of  two  parts :  1st.  The  Statement,  which  is  the  expressing  by  one 
or  more  equations  of  the  conditions  of  the  problem,  i.  e.,  the  rela- 
tions between  the  known  and  unknown  quantities  (parts  of  the 
figure)  to  be  compared;  and  2d.  The  Solution  of  these  equations, 
so  as  to  find  the  values  of  the  unknown  quantities  in  known  ones. 

810.  In  applying  the  equation  for  the  solution  of  such  problems 
as  are  now  proposed,  we  have  to  depend  upon  our  previously  ac- 
quired knowledge  of  the  properties  of  geometrical  figures  for  the 
relations  between  the  known  and  unknown  quantities,  which  will 
enable  us  to  form  the  necessary  equations,  i.  e.,  to  make  the  State- 


APPLICATIONS   OF  ALGEBRA   TO   GEOMETRY. 


277 


ment.  The  resolution  of  the  equations  thus  arising  is  effected  in 
the  ordinary  ways.  [See  Note,  page  239  of  The  Complete 
School  Algebra.] 

Sll.  The  details  of  this  method  will  be  most  readily  obtained 
from  a  careful  study  of  examples. 


EXAMPLES. 

812.  In  a  right  angled  triangle,  given  the  hypotenuse  and  the 
sum  of  the  other  two  sides,  to  find  these  sides  separately. 

A 

Solution. — Let  ABC  be  a  triangle,  right  angled 

at  B.  Let  the  known  hypotenuse  be  7i,  the  unknown 
base,  ?/;  the  unknown  altitude,  x;  and  the  knowns,\xvii 
of  the  base  and  altitude,  s. 

We  have  here  two  unknown  quantities,  and  hence 
must  have  two    equations,  in   order  to  find   their 
values.    One  of  these  equations  is  furnished  directly 
by  the  statement  of  the  problem,  which  says  that  the  sum  of  the  base  and  per- 
pendicular is  to  be  given.     Hence — 

Equation  1  is  x  +  y  =  s. 

A  second  relation  between  x  and  i/  and  the  known  quantit}'-  7i  is  furnished  by 
the  relation  given  in  Part  IL  [346).    Whence — 

Equation  2  is  x^  +  f=  ^^ 

Solving  these  equations  we  find — 

y  =  is±  ^^W^^\  and  x  —  is  T  W^^""  -  «'• 
If  7i  =  10  and  s  =  14,  we  find  x  —  G,  and  y  =  8 ;  or  .2;  =  8,  and  ?/  =  6. 

Geometrical  Solution — It  is  exceedingly  interesting  and  instructive  to 
compare  the  algebraic  solution  of  such  problems  with  their  geometrical  solution, 
when  the  problem  can  be  solved  in  both  ways.  The  geometrical  solution  of 
this  problem  is  as  follows : 

Take  DC  =  s,  the  sum  of  the  two  sides, 
and  make  ODC  .=  45°.  From  C  as  a  centre, 
with  a  radius  Ti,  the  hj^potenuse,  describe  an 
afc  cutting  DO,  as  in  A  and  A',  Draw  AC  and 
the  perpendicular  AB,  also  A'C  and  the  per- 
pcndicular  A'B'.  Both  the  triangles  ABC  and 
A'B'C  fulfilthe  conditions.  For  AB  =  DB  (?), 
whence  AB  +  BC  =  s,  and  AC  =  li,  by  con- 
struction. So,  also,  A'B'=  DB'  (?),  whence  A'B' 
+  B'C  =  8,  and  A'C  =  h,  by  construction. 


Comparisons  of  these  Solutions. — 1st. 
We  find   in  the  algebraic  solution,  that,  in 


Fig.  4:30. 


278  EXERCISES  IN  GEOMETEICAL  INVENTION. 


general,  y  may  have  two  values— viz.,  is  +  l^^h"^  —  «",  and  \s  —  ^^2,h^  —  s* ; 
and  that  when  y  =  U  +  ^/\/2/t^  —  s',  ;c  =  i«  —  |-y/2AT  —  .s" ;  but,  when  y  = 
^s  —  4^'^/2/t'^  —  A",  X  =  \s  4-  4-^/2/4^  —  s^.  Correspondingly,  we  find  in  the 
geometrical  solution  tlint  llie  base  (]/)  may  have  two  values — viz.,  BC,  and  B'C  ; 
and  that  when  the  base  is  BC,  the  altitude  {x)  is  AB;  but,  wiien  the  base  is 
B'C,  the  altitude  is  A'B'. 


2d.  From  the  algebraic  solution,  we  observe  that  the  base  y=.^s  ±  ^y  2/i^  —  »^ 
may  be  considered  as  made  up  of  two  parts — viz.,  a  rational  part,  \s,  and  a 
radical  part,  ^^2^  —  «» ;  and  that  the  altitude,  x  =  \s  ^  i^2?L^  -"7',  is  made 
up  of  the  same  parts,  only  observing  that,  if  the  base  is  considered  as  the  sum 
of  these  parts — viz.,  |s  +  i\/2h^  —  s^,  the  altitude  is  their  difference — viz. 
¥  —  iV^^i^  —  *'^-  If.  however,  the  base  is  is  —  iy\/2/iF  —  s\  the  altitude  is 
4.9  4.  iy^2h-  —  s^.  Now,  we  can  discover  exactly  the  same  things  geometrically, 
and  can  show  exactly  what  is  the  geometrical  meaning  of  each  of  the  parts  of 
the  values  of  y  and  x.  To  do  this,  draw  C/*  bisecting  AA' ;  let  fall  the  perpen- 
dicular/€,  and  draw  AA:  and  fg  parallel  to  DC.  C/is  perpendicular  to  DO  (?), 
and  hence  equal  to  D/  (?).     Also,  De  =  eC  =  fe  =  ^s  (?).    From  the  right  angled 

1  s 

isosceles  triangle  D/C,/C  =  —^  DC  =  —r    (?).      Hence,    from    A/C,    A/  =z 

^2  V2 

.  /ac'  -fQ^'  =  ^W  -  \s'  =  — ;i-v/^^^-^-    -A.gam,  from  the  right  angled 

r  y2 


isosceles  triangle  fiJcf,  we  have  A^  =  --:r  A/(?)  =  i^/2h^  -  s^.    But  Ak  =fk  = 

fg  =  A'g  =  Be  =  eS'.  Hence  we  see  that  the  rational  part  of  the  value  of  y  {^s) 
is  eC,  and  that  the  radical  part  {i^2?i-  —  s-)  is  Be,  or  eB'.  In  the  triangle  ABC 
the  81(771  of  these  parts  is  the  base;  and  in  the  triangle  A'B'C,  their  differe7ice  is 
the  base.  In  like  manner /^  represents  the  rational  part  of  the  value  of  x,  and 
fk  =  A'g,  the  radical  part. 

3d.  From  the  algebraic  solution  we  see  that  if  s-  =  2h'\  y  =  ^s,  and  x  =  ^s. 

The  same  thing  is  seen  in  the  geometrical  solution,  for  if  &^  =  2h'^,  h  =  — ps, 

or/C  ;  whence  the  arc  stinick  from  C  as  a  centre,  with  h  as  a  radius,  would  be 
tangent  to  DO,  instead  of  intersecting  it  in  two  points.  Again,  if  si^  >  2/t',  the 
quantity  under  the  radical  sign  is  negative,  and  the  radical  becomes  imagi7iary. 
This  means,  that  110  t7'iangle  can  be  formed  under  these  circumstances.    This 

case  appears  in  the  geometrical  solution  also,  for  then  h  <  —^s,  or  less  than 

\2 
/C,  and  consequently  the  arc  struck  from  C  as  a  centre,  with  radius  h,  will  not 
touch  DO,  and  we  get  no  triangle. 


*  This  part  of  the  construction  should  not  be  allowed  on  the  figure  till  it  is  wanted— i.e.,  till 
this  stage  of  the  discussion. 


APPLICATIONS  OF  ALGEBRA  TO   GEOMETRY.  279 

813.  ScH. — This  problem  is  discussed  thus  at  length  as  an  illustration  of 
what  may  be  done  by  such  methods.  Of  course,  all  problems  are  not  equally 
fruitful;  but  the  student  should  not  rest  satisfied  with  a  mere  determination  of 
the  values  of  the  unknown  parts  in  known  terms,  when  anything  farther  is 
revealed  either  by  the  process  or  result  of  the  algebraic  solution.  Especially 
should  he  desire  to  become  expert  in  seeing  what  geometrical  relations  are 
indicated  by  iha  form  of  the  answer  obtained. 


814:,  Given  the  lengths  of  the  medial  lines  from  the  acute  angles 
of  a  right  angled  triangle,  to  determine  the  triangle,  i.  e.,  to  find  the 
base  and  perpendicular. 

SuG's.— Let  AD  =  a,  CE  =  5,  AB  =  2x,  and  CB  =  2y;  then 

4i;2  +  2^2  ^  a\  and  4/  +  x''  =  b'{^).    .:  2x  =  AB  =3|/i^^lzii!, 

'  15 


and  2y  =  CB  =2^!^ 


A  E  B 


The  form  of  these  results  indicates  that  CB  sustains  the 
same  relation  to  CE  and  AD  that  AB  does  to  AD  and  CE — a  Fig.  431. 

fact  which  is  evident  from  the  nature  of  the  case. 

Again,  if  4rt"^  <  b'\2x^is  imaginary;  and  if  46-  <  a^,  2y  is  imaginary.  In 
either  case  the  triangle  cannot  exist.  So  also  if  4a^  =  ¥,  2x  =  0  /  and  if  46"^ 
=  a'\  2y  =  0,  and  there  can  be  no  triangle.  This  may  be  seen  from  the  figure 
by  conceiving  AB,  for  example,  to  diminish.  As  A  approaches  B,  AD  ap- 
proaches equality  with  DB,  and  CE  with  CB.     Hence  the  limit  is  AD  =  ^CE. 

Thus  we  see  thai  either  medial  line  must  be  more  than  half  tJie  other, — a  propo- 
sition which  is  proved  by  this  solution. 


815.   The  hypotenuse  and  radius  of  the  inscribed  circle  of  a 
right  angled  triangle  being  given,  to  determine  the  triangle. 

Results. — Calling  the  hypotenuse  7i,  the  radius  r,  the  base  x,  and  the  per- 
pendicular 3^,  we  have,  x 


_    2r    +    7i    ±     /y/   h?    -    4hr   - 

-   4r-' 

and 

2 

2r  +  hT  ^h?  -  4hr  -  Ar' 

y  = 

The  results  being  the  same  in  other  respects,  the  double  sign  before  the  radical 
indicates  that  the  base  and  perpendicular  are  interchangeable — a  flict  which  is 
evident  from  the  nature  of  the  case. 

If  the  radical  is  0,  i.  e.,  if  h^  —  4hr  —  ir^  =  0,  x  =  r  +  l\  and  y  =  r  +  ^h, 
and  the  base  and  perpendicular  are  equal.  Let  the  student  show  the  same  thing 
geometrically  (from  a  figure). 

Also,  if  ^2  _  4Jir  —  4;'2  =  0,  7i  =  2r  (1  ±  ^2).    In  this  result  the  negative 


280  EXERCISES  IN  GEOMETRICAL  INVENTION. 


sign  is  to  be  rejected,  since  it  would  make  h  negative,  as 
■x/2  >  1. 

The  value  U  =  2r  (1  +  ^2)  is  readily  seen  from  the  figure 
when  AB  =  CB.     Thus  AC  =  2DB  =  2  (DO  +  OB)  =  2  (r  + 

r  V^)  =  2/-  (1  +  V2)  (?). 


816.  A  tree  of  known  height  standing  perpen- 
dicular on  a  horizontal  plane,  breaks  so  that  its  top  strikes  the 
ground  at  a  given  distance  from  the  foot,  while  the  other  end  hangs 
on  the  stump.  How  high  is  the  stump  ?  That  is,  given  the  base 
and  the  sum  of  the  perpendicular  and  hyj^otenuse  of  a  right  angled 
triangle,  to  determine  the  perpendicular. 

Result. — Let  a  be  the  height  of  the  ti-ee,  &  the  distance  from  the  foot  to  the 
point  where  the  top  strikes,  and  x  the  height  of  the  stump  ;  then  x  = 
a'-  -  h^ 

2a     ' 

a-  —  b'  b^        b^ 

Since =  ^  —  —  ,    ^r-  is  the  distance  below  the  middle,  at  which 

2a  '  2a     2a 

the  tree  breaks. 


817.  In  a  rectangle,  knowing  the  diagonal  and  perimeter,  to  find 
the  sides. 


818.  Knowing  the  base,    b,  and  altitude,  «,  of  any  triangle,  to 
find  the  side  of  the  inscribed  square,  x. 

Result,  X  = r. 

a  -\-  b 


819.  In  an  equilateral  triangle,  given  the  lengths,  a,  b,  c,  of  the 
three  perpendiculars  from  a  point  within  upon  the  sides,  to  deter- 
mine the  sides. 

^g's. — Find  an  expression  for  the  altitude  in  terms  of  the  sides;  and  then 
get  two  expressions  for  the  area  of  the  whole  triangle.     Equate  these. 


Result,  each  side  = 


V3 


820.  In  a  right  angled  triangle  whose  hypotenuse  is  h,  and  differ- 
ence between  the  base  and  perpendicular  d,  to  find  these  sides. 

„      ,^             -  d±  \/W  -  (P        .     ,       d^  ^/2¥  -  d' 
Residts,  X  = :r ,  x  -\-  a  = ^ . 


APPLICATIONS   OF  ALGEBRA  TO   GEOMETRY.  281 

Queries. — Why  must  the  minus  sign  of  the  radical  be  omitted  in  the  geo- 
metrical interpretation  of  these  results  ?  What  is  the  least  possible  value  ofd? 
What  are  the  sides  of  the  triangle  for  these  values?  What  is  the  superior  limit 
of  the  value  of  d  ?    What  do  the  sides  become  at  this  limit  ? 


821,  In  an  equilateral  triangle  given  the  lines  a,  h,  c,  drawn  to  its 
three  vertices  from  a  point  within  or  without,  to  find  the  sides. 
Result. — Each  side  = 

i 


ft'  +  y  +  c'  d=  a/6  {a'b''  +  b'c'  +  cW)  -  3  (a'  +  b'  +  c*)  ) ^ 


The  radical  is  +  when  the  point  is  within,  and  —  when  it  is  without. 


822.  The  perimeter  of  a  right  angled  triangle  and  the  perpen- 
dicular from  the  right  angle  upon  the  hypotenuse  being  given,  to 
determine  the  triangle. 

Sug's. — Let  8  be  the  perimeter,  p  the  perpendicular  upon  the  hypotenuse,  and 
X  +  y,  X  —  y  the  two  sides  about  the  right  angle.  Then  the  hypotenuse  =  s 
—  2x,  and  we  readily  form  the  tAvo  equations  p  {s  —  2x)  =  x'^  —  y^,  and 

s(s  4-  2?)^ 

{x  +  yf  +  {x  —  yf  ={8  —  2xy  (?).    Hence  x  =  ^ —    and  this  value  substi- 
tuted in  either  equation  will  give  y. 


823.  The  base  of  a  plane  triangle  is  b  and  its  altitude  a,  required 
the  distance  from  the  vertex  at  which  a  parallel  to  the  base  must  cut 
the  altitude  in  order  to  bisect  the  triangle. 

Result,      —pr 
V2. 

Query. — What  does  the  fact  that  h  does  not  appear  in  the  result  show  ? 


824:.  Having  given  the  area  of  a  rectangle  inscribed  in  a  triangle, 
can  the  triangle  be  determined  .^  Can  it,  if  the  rectangle  is  a 
square  ?  If  the  rectangle  is  a  square  and  the  triangle  right  angled  ? 
If  the  rectangle  is  a  square  and  the  triangle  equilateral  ? 


825.  The  sides  of  a  triangle  being  «,  b,  c,  to  find  the  perpendicu- 
lar upon  c  from  the  opposite  angle. 


1 


Result,  2^  =  ^V2c'  {a'-\-b')  +  2a'b'  -  a*  -  b*  -  c\ 

Sug's. — Observe  that  a  and  b  are  similarly  involved  in  the  result,  but  c  is 
diffei-ently  involved  from  either.  This  is  evidently  as  it  should  be,  since  a  and 
b  are  the  sides  about  the  angle  from  which  ;>  is  let  fall ;  and  are  thus  similarly 
related  to  p.     But  c,  the  side  on  which  p  falls,  is  differently  related  to  p  from 


282  EXERCISES  IN  GEOMETRICAL  INVENTION. 

either  of  the  others.    The  student  should  be  able  to  write  the  value  of  the  per- 
pendiculars upon  each  of  the  other  sides,  from  this  one.    Thus,  that  on  a  is 


826.  The  sides  of  a  triangle  are  a,  h,  c,  to  find  the  side  of  an  in- 
scribed square  one  of  whose  sides  falls  in  c. 

Sug's. — The  altitude  may  be  found  from  the  preceding,  hence  may  be  as- 
sumed  as  known.     Call  it  p.    Then  the  side  of  the  requu'ed  squai-e  is 


c  +  p 
What  is  the  side  of  the  square  standing  on  a?    On  6 ? 

Query. — Will  the  square  be  the  same  on  whichever  side  it  stands  ?  Observe 
that  though  the  values  here  found  are  apparently  different,  they  may  not  be  so 
really,  since  p  is  different  in  each  case.    But  let  the  student  decide. 


827'  Having  the  area  of  a  rectangle  inscribed  in  a  given  triangle 
and  standing  on  a  specified  side,  to  determine  the  sides  of  the 
rectangle. 

Besult,  h  being  the  base  on  which  the  rectangle  stands,  j9  the  alti- 
tude from  this  base,  and  s  the  given  area,  we  have  for  the  sides 

h  /W     sb        ,  p         .  ///       sp 

^  =  3  ^|/ 1  -  ?  ^'"^  ^  =  2  ^|/i  -y 

Sro's. — The  ±  and  t  signs  indicate  that,  in  general,  there  can  be  two  equal 
rectangles  inscribed  standing  on  the  same  base.  The  student  will  do  well  to 
illustrate  it  with  definite  numerical  values,  as  p  =  10,  i  =  G,  s  =  10. 

Again,  —  must  be  greater  than  — ,  J^nd  j-  >  '^,  i.  e.,  s  must  be  less  than  ^pb. 

That  is,  the  greatest  rectangle  is  half  the  area  of  the  triangle,  since  i  ^  is  the 
area  of  the  triangle. 


828.  The  Algebraic  solution  of  a  problem  often  enables  us  to 
effect  a  geometrical  construction.     We  will  give  a  few  examples. 

Through  a  given  point  within  a  circle,  to  draw  a 
chord  of  a  given  length. 

SoLUTiox.— Let  8  be  the  length  of  the  required  chord,  and 
P  the  given  point.  Since  P  is  a  known  point,  call  AP 
=  a,  PB  =  b,  AB  being  the  diameter  througJi  P.  Let  CD 
represent  the  required  chord,  and  calliug  CP,  r,  PD  =  s  —  x. 
Then  sx  —  x-  —  ah ;  whence  x  —  Isi  ±  Vi*  —  ''<^- 


Fig.  4:33. 


APPLICATIONS   OF  ALGEBRA  TO   GEOMETRY. 


283 


To  effect  the  geometrical  construction,  let  8  be  the  length 
of  the  given  chord,  and  P  the  point  in  the  given  circle.  Draw 
the  diameter  through  P,  and  erect  PE  perpendicular  to  it. 
Make  EH  =  is;  then  since  PE^  =:  ab,  PH  =  -y/i  s^  —  ab. 
Now  take  H\  =  is,  and  from  P  as  a  centre,  with  a  radius  PI 


=  is  +  /^i«'^  —  ab,  strike  the  arc  Dl  intersecting  the  circum- 
ference.    DPC  is  the  chord  reqmred. 


Fig.  4S4. 


From  the  radical  Vi*"  ~  ^^^  we  see  that,  if  ab  >  ^s-,  x  is 
imaginary,  as  we  say  in  algebra.     In  such  a  case  the  problem  is  geometrically 
impossible,  as  will  appear  from  the  construction,  for  then  PE  is  greater  than 


EH,  which  makes  HP,  the  representative  of  -^/l^^ 
X  has  but  one  value,  and  the  segments  are  equal. 


ab,  impossible.    If  \8'^  =  ab^ 


829.  To  find  a  point  in  a  tangent  to  a  circle  from  which,  if  a 
secant  be  drawn  to  the  extremity  of  the  diameter  passing  through  the 
point  of  tangency,  the  external  segment  shall  have  a  given  length. 


Solution. — Let  AB  =  d  he  the  diameter  of  the 
given  circle,  DX  =  a  the  external  segment  of  the  re- 
quired secant,  and  the  whole  secant  BX  =  x.  Then 
x'^  —  ax  =  cP,  and  x  =  ia  ±  ■\/lV  +  \a;K 

To  effect  the  geometrical  construction,  construct 
the  radical  by  taking  AC  =  \a\  whence  BC  =: 
■yjd^  +  \cC\  Now  make  CY  =  \a,  and  with  B  as  a 
centre,  and  BY  as  a  radius,  strike  an  arc  cutting  the 
tangent,  as  in  X.     Then  is 


Fig.  435. 


BX^.x  =  ia+  ^/cP  +  \a\ 

The  negative  value  of  the  radical  is  inapplicable  in  this  elementary,  geomet- 
rical sense,  since  as  -y/cP  +  W  >  i«>  this  would  make  x  a  negative  quantity. 
Again  we  see  that  no  real  value  of  a  can  render  x  imaginary. 

We  can  observe  the  same  things  from  the  geometrical  construction.  Thus, 
if  the  negative  value  of  the  radical  were  taken,  x  would  be  numencally  less 
than  BC,  by  ia,.or  AC.  But  BC  —  AC  <  BA.  Hence  an  arc  struck  from  B 
with  the  required  radius  would  not  cut  the  tangent.  "We  see  also  that  a  may 
have  any  value  between  0  and  oo . 


8S0.  Given  the  hypotenuse  and  area  of  a  right  angled  triangle, 
to  construct  the  triangle. 


Sug's. — Let  li  be  the  hypotenuse,  s"^  the  area,  and  x  the 
perpendicuhir  from  the  right  angle  upon  the  hypotenuse. 
Then  hx  =  2s'^,  or  ih  :  s  : :  s  :  x,  and  7i  :  2s  :  :  2s  :  2x. 

The  figure  will  suggest  the  construction. 


284 


EXERCISES  IN  GEOMETRICAL  INVENTION. 


831.  Through  a  point  between  two  lines  which  intersect,  to  draw 
a  line  which  shall  cut  off  a  triangle  of  given  area. 

Sug's. — Let  AY  =  x,  and  the  requhed  area  =  «'. 

We  have  h  :H  ::  x  -  h  :  x.     .'.   H   =  -^. 

X  —  0 


And  Hx  =  2sK     .:  H 


2*2 
X  ' 


Thus 


=  ^-/K^4 


To  construct  this,  find  c  =  ^,  i,  e.,  construct 


a  third  proportional  to  h  and  ."*.  Then  construct  yc  (c  —  26),  i.  e.,  find  a  mean 
proportional  between  c  and  c  —  26 ;  let  this  be  m.  Whence  x  =  c  ■±:  m.  In  gen- 
eral, there  may  be  two  solutions,  if  any,  since  there  are  two  values  of  re.     [This. 

should  also  be  observed  from  the  figure.]    But  if  26  >  t  there  is  no  solution. 


If  y   =  26,  there  is  but  one  solution.    In  the  latter  case  where  is  the  given  point 

0  ?    What  is  the  geometrical  difl3culty  when  26  >  -  ?  Can  m  be  numerically 

III 

greater  than  c  ? 


832.  To  construct  the  four  forms  of  the  affected  or  complete 


quadratic  equation,  viz.,  (1.) 


0,  (3.)  x'  -  px  +  q 
equations. 


0,  (4.) 


+  2)X  —  q  =  0,  (2.)  X*  —  px  —  q  = 
'^  +  px  -f  (^  =  0,  without  solving  the 


First  Form:,  x-  +  px  —  q  =  0. — Draw  any 
two  lines  as  LM,  NP,  intersecting  in  some  point 
O.  Resolve  q  of  the  equation  into  two  factors, 
as  r  and  q\  so  that  we  have  x'^  +  px  —  r  x  q'  = 
0.  Take  OA  =  ;?,  OB  =  r,  OC  =  q'.  Bisect  CB 
and  AO  by  perpendiculars,  and  from  their  in- 
tersection F  as  a  centre,  with  a  radius  FB,  draw 
a  circle.  Then  DO,  or  AE,  is  x,  the  positive 
root.  For  x  {x  +  p)  =  rq',  or  x"^  +  px  —  rq'  =  0. 
The  negative  root  is  OE.  Thus,  let  OE  =  (—  x). 
Theu  DO  =  AE  =  (-  a;  -  p).    Hence  (-  x) 


^0. 
This  construction  is  evidently  always  possible  irrespective  of  the  relative 
magnitudes  of  p,  r,q' ;  a  fact  which  agrees  with  the  statement  in  algebra  that 
this  form  always  has  real  roots. 

Second  Form,  x^  —px  —  rq'  =  0. — The  construction  is  the  same  as  for  the 
first  form;  only,  in  this  case  OE  is  the  positive,  and  DO  the  negative  root. 
Thus  for  OE  =  x  (positive),  we  have  DO  x  OE  =  (x  —  p)  x  =  rq',  or  x^  —  px  — 


APPLICATIONS    OF   ALGEBRA   TO    GEOMETRY. 


285 


rg'  =  0.    For  DO  =  (-  x),  we  have  DO  x  OE  =  DO  (OA  +  AE)  =  DO  (OA  +  DO) 
=  (—  -i)  {p  —  ^)  =  ^'Q\  01'  -^"^  —  P-^  —  ^'?'  =  0. 

Observe  that  in  the  first  case  the  negative  root  is  numericall}'  greater  than 
the  positive  ;  while  it  is  the  reverse  in  tliis  form.  This  agrees  with  tlie  conclu- 
sions of  algebra  (See  Complete  School  Algebra,  104). 

Thikd  Form,  x-  —  px  +  rq'  =0. — 
Draw  any  two  lines,  as  OM,  OP,  meet- 
ing at  0.  Take  OA  =  ;?,  OB  =  r  or  q', 
and  OC  =  q'  or  r.  Erect  perpendiculars 
at  the  middle  points  of  OA,  and  BC  ; 
and  from  their  intersection  F  as  a  cen- 
tre, with  a  radius  FB,  strike  a  circum- 
ference. Then  OE  and  OD  are  the 
values  of  x.  For  OE  =  .t,  OE  x  OD  = 
OE  X  EA  =  OE  (OA  -OE.)  =  x{p-  .?) 
=  rq\  or  x'^  —  px  +  rq'  =  0.  For  OD 
=  X,  OD  X  OE  =  OD  (OA  -  AE)  =  OD  (OA 
—  px  +  7'q'  =  0. 

Observe  that  the  former  value  of  x  is  greater  than  the  latter,  but  that  neither 
is  negative. 

So  also,  we  may  readily  see  that  the  roots  ma)-  become  equal,  and  also,  im- 
aginary. Tlius  if  the  circle  were  tangent  to  OA,  the  roots  would  be  equal,  and 
if  it  did  not  touch  OA  they  would  both  be  imaginary.  (See  Algebra,  as 
above.) 


Fig.  439. 


OD)  =  X  ip  —  x)  =  rq\  or  a;- 


FouRTH  Form,    x'^  +  px  +  rq'  =  0.— The  constniction  is  the  same  as  the 


last,  only  both  values  of  x  are  negative.     Thus,  (—  x)  [p 
(p  +  x)  =  rq',  —  px  —  X-  —  7'q'  =  0,  or  x-  +  px  +  rq'  =  0. 


(-  ^0]  =  (-  ^O 


ScH. — Thus  we  see  that  we  can  construct  any  equation  of  the  second  degree 
containing  but  one  unknown  quantity,  which  has  real  roots.  Hence,  if  the  al- 
gebraic solution  of  a  geometrical  problem  requires  only  the  resolution  of  such  an 
equation,  the  algebraic  solution  will  lead  to  the  geometrical  construction. 


833.  "We  have  now  given  sufficient  illustrations  of  this  most  in- 
teresting and  important  subject,  so  that  the  student  should  have 
caught  the  spirit  of  this  method  of  using  algebra  to  subserve  the 
•^iurposes  of  geometrical  investigation.  We  shall  simply  append  a 
list  of  problems,  upon  which  the  student  can  put  in  exercise  both 
his  algebraic  and  geometric  knowledge.  But  we  cannot  refrain 
from  repeating  the  advice,  that  the  learner  should  not  rest  satisfied 
with  the  mere  algebraic  resolution  of  the  problem.  ITc  should  be 
ambitious  to  trace,  as  fully  as  possible,  the  wonderful  relations  which 
exist  between  the  abstract  operations  of  algebra,  and  *he  more  con- 
crete relations  of  geometry. 


286  EXERCISES  IN  GEOMETRICAL  INVENTION. 


EXA3IPLES. 

834:.  Given  the  perimeter  of  a  right  angled  tri- 
y     angle  and  the  radius  of  the  inscribed  circle,  to  de- 


termine the  triangle. 


835.  Given  the  hypotenuse  of  a  right  angled 
triangle  and  the  side  of  the  inscribed  square,  to  de- 
termine the  triano:le. 


836.  In  a  right  angled  triangle,  given  the  radius  of  the  inscribed 
circle,  and  the  side  of  the  inscribed  square,  the  right  angle  of  the 
triangle  constituting  one  angle  of  the  square,  to  determine  the 
triangle. 

Sug's. — Letting  x  and  y  be  the  sides,  z  the  hypotenuse,  r  the  radius  of  the 

inscribed  circle,  and  s  the  side  of  the  inscribed  square,  we  have  «  =  — ^—* 

X  -\-y 

xy  =  r  {x  +  y  +  z),  and  x  +  y  =  z  +  2r.  TThence  z  =  2r  l~ 1,  etc 


837'  In  any  triangle  whose  sides  are  «,  J,  c,  to  find  the  radius  of 
the  inscribed  circle. 


838.  Show  that  the  area  of  a  regular  dodecagon  inscribed  in  a 
circle  whose  radius  is  1,  is  3. 


839.  Find    the    area    of    a   regular  octagon 
a\;®*^     whose  side  is  a. 

Result,  2  (V2  4-  1)  a\ 


840.  Find  the  radii  of  three  equal  circles  de- 
FiG.  441.  scribed  in  a  given  circle,  tangent  to  the  given 

circle  and  to  each  other. 


841.  The  space  between  three  equal  circles  tangent  to  each  other 
is  a ;  what  is  the  radius  ? 


842.  In  a  triaugle,  given  the  ratio  of  two  sides,  and  the  segments 
of  the  third  side  made  by  a  perpendicular  let  fall  from  the  angle 
opposite. 


APPLICATIONS   OF   ALGEBRA   TO    GEOMETRY.  287 

843'  In  a  triangle,  given  the  base  and  altitude,  and  the  ratio  of 
the  other  sides,  to  determine  the  triangle. 

844'  Given  the  base,  the  medial  line,  and  the  sum  of  the  other 
sides  of  a  triangle,  to  determine  the  triangle. 

84S.  To  determine  a  right  angled  triangle,  knowing  the  perim- 
eter and  area. 

Sug's.  x"^  +  y^  =  z\  X  +  y  +  z  =  2p,  and  xy  =  2s\  give  y  +  x  =  2p  —  '^, 
x'^  +  2xy  +  y^  =  4p"^  —  4pz    +    2^,    2^    +    4^-2  —  ^f  —  4pz  +  z'^ ;    whence 

z  =  ' .    Now  use  y  +  a;  =  2»  —  z  =  ,  and  xy  =  2s^. 

p  if  1^  P     '  "^ 


846.  To  determine  a  right  angled  triangle,  knowing  the  perim- 
eter, and  the  sum  of  the  hypotenuse,  and  the  perpendicular  upon  the 
hypotenuse  from  the  right  angle. 

Sug's.  x"^  -^  y'^  =  z'^,  x  -^  y  -{■  z  =  2p,  z  +  u  =  a,  xy  =  zu.  Then 
a;"^  +  2xy  +  y'^  =  4:p'^  —  4pz  +  s^ ;  whence  2xy  =  4p'^  —  Apz,  and  hence 
2s  (a  —  2)  =  4p  —  4^)2,  etc. 


847'  The  volume,  the  altitude,  and  a  side  of  one  of  the  bases  of 
the  frustum  of  a  square  pyramid  being  known,  to  determine  a  side 
of  the  other  base. 


848.  To  determine  a  right  angled  triangle,  knowing  the  perim- 
eter, and  the  perpendicular  let  fall  from  the  right  angle  upon  the 
hypotenuse. 


849.  To  determine  a  triangle,  knowing  the  base,  the  altitude, 
and  the  difference  of  the  other  sides. 


8S0'  To   determine  a  triangle,  knowing  the  base,  the  altitude, 
and  the  rectangle  of  the  other  sides. 


851'  To  determine  a  right  angled  triangle,  knowing  the  hypote- 
nuse and  the  difference  between  the  lines  drawn  from  the  acute 
amgles  to  the  centre  of  the  inscribed  circle. 

Sug's.— Let  fall  CD  a  perpendicular  upon  AO  produced. 
Now,  since  the  the  angles  BAG  and  ACB  are  bisected, 
and  COD  =  QAC  +  OCA,  and  ICD  =  lAB,  they  being 
complements  of  the  equal  angles  CID,  lAB,  we  have,  COD 
=  OCD,  and  CD  =  OD  =  -y/^  CO.  Hence,  putting  AC 
=  A,  CO  =  X,  and  ^O  =  x  +  d,  we  have 

{x   +  d    +    ^i  xy  +  (-y/^  x)'   =  h^.    From  this  a;  is  readily  found. 
The  student  should  then  be  able  to  complete  the  solution.  / 


288 


INTRODUCTION   TO  MODERN  GEOMETRY. 


852.  Given  two  sides  of  a  triangle  and  the  bisector  of  their  in- 
cluded angle,  to  determine  the  triangle. 

853.  Given  the  three  medial  lines,  to  determine  a  triangle. 


854:.  Given  the  three  sides  of  a  triangle,  to  determine  the  radius 
of  the  circumscribed  circle. 


855.  Four  equal  balls  whose  radius  is  r  are  placed  on  a  plane  so 
that  each  is  tangent  to  the  other  three,  thns  forming  a  pyi'amid; 
what  is  its  altitude  ? 


856.  Given  the  base  of  a  triangle,  the  bisector  of  the  opposite 
ano-le,  and  the  radius  of  the  circumscribing  circle,  to  determine  the 


Sug's.— First  to  fiud  ED  =  x.    Since 


EM  =  r  —  -y^r^  —  b'\  it  may  be  considered  known  and 

put  equal  to  c.  We  then  have  DM  =  -^/c^  +  x'^ ;  and  also, 

59  _  3.2 
DM  X  a  :=:  AD  X    DB   =  ^/-^  —  x-,  or  DM  =    . 

Whence  Vc^  +  £^  =   '  ~  ^  ,  and  x  is  readily  found. 

Calling  ED  =  s,  the  student  will  have  no  difficulty  in 
proceeding  with  the  solution. 


CHAPTEE  11.   . 

INTBOnUCTION  TO  MODERN  GEOMETRY.* 


SECTION  I. 


OF  LOCI. 

857.  The  term  Locus  t,  as  used  in  geometry,  is  nearly  synony- 
mous with  geometrical  figure,  yet  having  a  latitude  in  its  use  which 
the  other  does  not  possess.     The  locus  of  a  point  is  the  line  (geo- 

*  With  strict  propriety  only  the  latter  sections  of  this  chapter  helong  to  the  Modem  Geome- 
tnj.  technically  so  called.  Bnt,  as  the  entire  chapiter  is  composed  of  matter  which  has  not 
hitherto  found  place  in  our  common  text-books,  and  the  relative  importance  of  which  is  be- 
coming more  fully  appreciated  in  modern  times,  the  author  has  ventured  to  embrace  the  whole 
under  this  title. 

t  The  word  Locus  is  the  Latin  tor  place. 


OF  LOCI.  289 

metrical  figure)  generated  by  the  motion  of  the  point  according  to 
some  given  law. 

In  the  same  manner,  a  surface  is  conceived  as  the  locus  of  a  line 
moving  in  some  determinate  manner. 

Ill's.— The  locus  of  a  point  in  a  plane,  which  point  is  always  equidistant 
from  the  extremities  of  a  given  right  line,  is  a  straight 
line  perpendicular  to  the  given  line  at  its  middle  point.  C 
Thus,  suppose  AB  a  fixed  line,  and  tlie  locus  of  a  point 
equidistant  from  its  extremities  is  required ;  that  point 
may  be  anywhere  in  a  perpendicular  to  AB  at  its  mid- 
dle point,  and  cannot  he  anywhere  else  in  this  plane.  a 

This  perpendicular  is  the  locus  (place)  of  a  point 
subject  to  the  given  law. 

Again,  a  boy  on  the  green  is  required  to  keep  at  just  ^ 

20  feet  from  a  certain  stake  ;  where  may  he  be  found?  P^^  ^^ 

i.  e.,  what  is  his  locus  (place)  ?    Evidently,  the  circum- 
ference of  a  circle  whose  radius  is  20  feet.  Thus,  the  locus  of  a  point  in  a  plane, 
equidistant  from  a  given  point,  is  the  circumference  of  a  circle.    This  is  the 
place  of  such  a  point. 

What  is  the  locus  in  space  of  a  point  equidistant  from  a  given  point  ? 

"What  is  the  locus  of  a  point  in  space  equidistant  from  the  extremities  of  a 
given  line  ?    A  plane. 

What  is  the  locus  of  a  line  moving  so  that  each  point  in  it  traces  a  right  line  ? 
In  general,  a  plane  ;  if  it  move  in  the  direction  of  its  length,  a  straight  line. 

What  is  the  locus  of  a  right  line  parallel  to  and  equidistant  from  a  given  line? 

What  is  the  locus  of  a  right  line  intersecting  a  given  line  at  a  constant 
angle  ?  *    A  conical  surface  of  revolution. 

What  is  the  locus  of  a  semicircle  revolving  on  its  diameter  ? 


PROPOSITIOIVS  AXD  PROBLEIIS  Ii\  DETERMIXIXG  PLANE  LO€L 

[Note.— The  student  should  be  required  to  give  every  demonstration  in  form, 
and  in  detail.  Frequent  exercise  in  writing  out  demonstrations,  is  almost  the 
only  method  of  securing  a  good,  independent  style  in  demonstration] 

858.  Theo. — The  locus  of  a 2:)oint  in  a  plane,  equidistant  from 
the  extremities  of  a  given  liiie,  is  a per2)encUcular  to  that  line  at  its 
middle  point. 

SuG. — To  prove  tiiis  we  have  simply  to  show  that  every  point  in  such  a  per- 
pendicular is  equidistant  from  the  extremities  of  the  given  line,  and  that  no 
other  point  has  this  property  (Part  II.,  129). 

*  That  is,  an  angle  which  remains  of  the  same  size. 

19 


290  INTRODUCTION  TO   MODERN  GEOMETEY. 

SS9.  JProb. — Fi7id  the  locus  of  a  jjoint  at  atiy  constant  distance 
m  from  a  straight  line.  Of  what  proposition  in  Part  II.  is  this  the 
converse  ? 

Sug's. — To  prove  the  proposition  which  the  answer  to  this  question  asserts,  it 

will  be  necessary  to  show  that  every  point  in  the  affirmed  locus  is  at  the  same 

distance  from  the  given  line  and  that  no  other  point 

is  at  that  distance.     We  aflBrm  that  the  locus  is  tieo 

right  lines  parallel  to  the  given  line  and  at  a  distance 

m  therefrom.    The  formal  demonstration  is  as  follows : 

Let  AB  be  the  given  line,  and  OE,  OE',  perpendiculars 

thereto,  each  equal  to  m.     Through  E  and    E'  draw 

CD  and  CD'  parallel   to  AB ;    then  is  CD,  CD',  the 

Fig.  445.  locus  required.*    For,  by  Part  II.  {156),  every  point 

in  CD,  CD',  is  at  the  distance  m  from  AB  ;  and  we  may 

readily  show  that  any  other  point,  as  P  or  P',  is  at  a  distance  greater  or  less  than 

m  fromAB.    Hence  CD,  CD',  is  the  locus  required. 


P 

E 

A 

P' 

:/n 

Oim 

C 

t:' 

SGO.  Theo. — In  a  circle,  the  locus  of  the  centre  of  a  chord  ^jar- 
allel  to  a  given  line  is  a  diameter. 

Dem. — Let  mn  be  any  circle,  and  AB  a  given  line. 
Then  is  the  locus  of  the  centre  of  a  chord  parallel  to  AB, 
a  diameter  of  the  circle. 

For,  let  DH  be  any  chord  parallel  to  AB.  Through  the 
centre  of  tlie  circle  C,  and  P,  the  middle  point  of  DH, 
draw  EL.  Now  EL  is  perpendicular  to  DH  (?),  and  con- 
sequently to  AB  (?).  Then  will  EL  be  perpendicular  to 
any  and  eveiy  chord  parallel  to  DH  (?),  and  hence  will 
bisect  such  chord  (?).  Therefore  the  locus  of  the  centre  of  a  chord  parallel  to 
AB  is  a  diameter. 

Again,  any  point  in  the  circle  and  out  of  the  line  EL  is  not  the  middle  point 
of  chord  parallel  to  AB.  Thus,  letting  P'  be  such  a  point,  draw  a  chord 
through  P'  parallelto  AB.  As  there  can  be  but  one  such  chord  (V),  and  as  EL 
bisects  it  (?),  P'  is  without  the  diameter  (?). 


Fig.  446. 


861.  Tlieo. — Tlie  locus  of  the  centre  of  a  circumference  passing 
through  tiuo  given  poi)its  is  a  straight  line. 

SuG.— Consult  Part  tl.  {159,  168,  197),  The  student  should  put  the 
argument  in  form. 

862.  Tlieo. — The  locus  of  the  centre  of  a  circle  lohich  is  tan- 

*  It  is  important  that  the  student  think  of  these  two  lines  a?  one  locns,  or  as  parts  of  one  and 
the  same  locus,  if  this  will  aid  the  conception.  A  locus  may  consist  of  any  number  of  detached 
parts;  all  that  is  necessary  being  that  the  given  conditions  be  fulfilled.  In  this  respect  the 
word  locus  has  a  more  enlarged  meaning  than  the  term  (jeoineh^ai figure. 


OP  LOCI. 


291 


gent  to  a  given  circle  at  a  given  j)oint,  is  a  straight  line  passing 
through  the  centre  of  the  given  circle. 

Dem. — Let  C  be  the  centre  of  thegiveu  circle, 
and  B  the  point  in  the  circumference  to  which 
the  circle  *  shall  be  tangent,  the  locus  of  whose 
centre  is  required.  Through  B  drawTL  tangent 
to  the  given  circle.  Now,  a  circle  passing  through 
B,  and  tangent  to  the  given  circle,  will  have  TL 
for  its  tangent  (?),  and  as  a  radius  is  perpendicu- 
lar to  a  tangent  at  its  extremity,  and  only  one 
perpendicular  can  be  drawn  to  TL  through  B, 

the  centre  of  a  circle  tangent  to  the  given  circle  at  B  must  be  in  this  straight 
line.  Moreover,  as  the  given  circle  is  tangent  to  the  right  line  TL  at  B,  its 
centre  is  in  the  perpendicular  AX.     Hence  AX  is  the  locus  required. 


863'  Theo. — The  locus  of  the  centre  of  a  circle  of  given  radius 
R,  and  tangent  to  a  given  straight  line,  is  two  parallels  to  this  line  at 
a  distance  R  therefrom,  on  each  side.     Give  proof  in  form. 


864.  JProb. — Fi7id  the  locus  of  the  centre  of  a  circle  of  given 
radius  R,  whose  circumference  ptdsses  through  a  given  point.  Give 
proof  in  form. 


86^.  Tlieo. — The  locus  of  the  centre  of  a  lirie  of  constant  length, 
having  its  extremities  in  two  fixed  lines  luhich  cut  each  other  at  right 
angles,  is  the  circimifere?ice  of  a  circle. 

Sug's.— Let  MN  be  the  length  of  the 
given  line,  and  CD,  and  AB,  the  two  lines 
intersecting  at  right  angles,  in  which  the 
extremities  of  MN  are  to  remain.  Now, 
in  whatever  position  MN  may  be  placed, 
its  middle  point,  P,  is  at  the  same  distance 
(|MN)  from  O  (?).  To  show  that  any  point 
not  in  this  circumference,  as  <p,  is  not  the 
middle  point  of  a  line  equal  to  MN  passing 
Hirough  it,  and  limited  by  the  fixed  lines, 
from  (pas  a  centre,  with  a  radius  ^MN  cut 
CD,  as  in  C  ;  and  from  C  as  a  centre  with 
the  same  radius  strike  the  arc  ^P.  If  <p  is  without  the  circle,  CB  >  MN,  if 
Avithiu,  less.  Hence,  the  required  locus  is  a  circumference  whose  centre  is  0, 
and  whose  radius  is  iMN. 


Fig.  44S. 


*  Observe  the  form  of  expression.  We  say  "  tlie  circle,"  and  not.  "the  circles."  usin:;  the 
term  in  a  generic  sense,  as  including  all  which  have  the  required  property,  i.  e.,  all  which  arc 
tangent  to  the  given  circle  at  B. 


292 


INTRODUCTION  TO  MODEEN  GEOMETEY. 


866'  JProh. — Find  the  locus  of  the  centre  of  a  chord  of  constant 
length,  in  a  given  circle. 


itl 


SuG. — We  say,  once  again,  always  give  the  proof  in  form. 


867.  I^roh. — Find  the  locus  of  the  vertex  of  the  right  angle  of  a 
right  angled  triaiigle  of  a  constant  hypotenuse. 


868.  JProb. — Find  the  locus  of  the  middle  poi7it  of  the  chord  in- 
'  tercepted  on  a  line  through  a  given  point,  hy  a  given  circumference, 
when  the  given  point  is  without  the  circumference,  when  it  is  in, 
■    and  when  it  is  within  the  circumference. 


869.  JProh.— Find  the  locus  of  a  point  the  sum  of  whose  dis- 
tances from  ttoo  fixed  intersecting  liiies  is  constant,  i.  e.,  is  equal  to 
a  given  line. 

Solution.— Let  AB  and  CE  be  the  fixed 
lines,  and  m  the  constant  distance.  Draw 
MN  parallel  to  AB,  and  at  a  distance  m 
^  from  it.  Bisect  the  angle  CPN.  Then  is 
LR  (a  part  of)  the  locus  required.  For 
[the  student  will  here  show  that  the  sum 
of  the  distances  from  any  point  in  LR  to 
AB  aud  CE,  as  PD,  P"D"  +  P"d",  P'D' 
+  P'd',  P"'d"',  P'^D''  +  P'V^  is  con- 
stant and  equal  to  w],  observe  that  when 
one  of  the  perpendiculars  measuring  the  distance  from  a  point  in  the  locus, 
changes  from  one  side  to  the  other  of  the  line  on  which  it  is  let  fall,  its  sign 
changes.  Thus  P"D",  P"d"  being  considered  +,  P'd'  and  P"D''  are  to  be  con- 
sidered — .  This  is  a  general  principal  in  mathematics.  See  Part  II .  {215), 
and  foot  note. 

Finally,  LR  is  only  a  part  of  the  locus,  since  there  is  another  line  on  the  op- 
posite side  of  AB,  obtained  by  drawing  the  auxiliary  MN  on  that  side,  which 
fulfills  the  same  condition.  The  student  should  show  what  the  result  is  when  we 
draw  the  auxiliary  parallel  to  CE,  and  on  either  side  of  it,  also  that  any  point 
not  in  oue  of  these  lines  cannot  fulfill  the  required  condition.  The  complete 
locus  is  four  indefinite  right  lines  intersecting  each  other  at  right  angles,  so  as 
to  inclose  a  rectangle. 


Fig.  449. 


870.  I*voh. — Find  the  locus  of  a  point  such  that  the  sum  of 
the  squares  of  its  distances  from  two  fixed  points  shall  he  equivalent 
to  the  sqtiare  of  the  distance  between  the  fixed  ptoints. 


871-  JProb. — Find  the  locus  of  the  intersection  of  two  secants 


OF  LOCI. 


293 


drawn  through  the  extremities  of  a  fixed 
diameter  in  a  given  circle,  one  of  the  secants 
heing  alivays  perpendicular  to  a  tangent  to 
the  circle  at  the  point  where  the  other 
cuts  it. 


Sug's.  P  being  the  point,  show  that  PB  =  AB, 
for  any  position  of  AP  and  BP.  Hence,  any  point 
in  the  circumference  having  B  for  its  centre,  and 
AB  for  its  radius,  fulfills  the  conditions.  Show  that 
any  point  out  of  the  ciroumference  does  not  fulfill  the 
conditions. 


872.  JProb. — Find  the  locus  of  the  infrrsection  of  tioo   lines 
draimi   from  the  acute  angles    of  a  right 
angled  triangle,  through  the  points  vjhere  the 
po'pendicular  to  the  hgjjotciiuse  cuts  the  op-  /\  ■.    \b 

posite  sides,  or  sides  produced. 


Sug's. — The  locus  of  P  is  required.  Prove  that 
APC  is  always  a  right  angled  triangle,  wherever  the 
perpendicular  EF  to  the  liypoteuuse  AC  is  drawn. 


873.  JPvoh. — Find  the  locus  of  a  point  luhich  divides  a  line 
draionfrom  a  fixed  point  to  a  fixed  line  i?i  a  fixed  ratio. 

Sug's. — Most  problems  in  finding  loci  such  as  are  treated  in  Elementary 
Plane  Geometry,  viz.,  right  lines  and  circles,  are  readily  solved 
by  constructing  a  few  points  according  to  the  given  conditions, 
whence  we  can  determine  by  inspection  whether  the  required 
locus  is  a  right  line  or  the  circumference  of  a  circle;  and, 
having  discovered  this  fact  by  inspection,  it  will  remain  to 
show  why  it  sliould  be  so.  Thus,  in  the  present  problem,  O 
being  the  fixed. point,  and  AB  the  fixed  line,  drawing  a  few 
lines,  OC,  according  to  the  requirements,  and  dividing  them 
i^  the  same  ratio  (in  the  figure  3:2),  we  find  a  few  points 
P  in  the  locus.  We  then  discover  at  once  that  the  locus  is 
a  right  line  parallel  to  AB,  and  can  easily  see  w7iy  it  should 
be  so. 


Fig.  452. 


874.  I^rob. —  Two  fixed  circumferences  intersect:  to  find  the 
locus  of  the  middle  point  of  the  line  drawn  through  one  of  the  points 
of  intersection  and  terminated  ly  its  other  intersections  ivith  the  cir- 
cumferences. 


294 


INTRODUCTION   TO   MODERN  GEOMETRY. 


FUr.     45:3. 


Bug's. — TTe  will  first  give  an  example  of  the  course  TNhicli  the  miud  of  the 
student  might  take  iu  his  eflforts  to  discover  tlie  solution.    He  would  naturally 

draw  two  iinequal  *  circles,  as  M  and  N, 
and,  through  one  of  the  points  of  intersec- 
tion, as  A,  draw  BC,and  bisect  it  at  P.  It  is 
the  locus  of  P  that  is  desh-ed.  Now,  sup- 
pose the  line  BC  to  revolve  about  A,  B 
passing  towards  B',  and  C  towards  A.  It  is 
the  path  of  the  middle  point  that  he  seeks. 
When  C  reaches  A,  the  line  becomes  tangent 
to  N,  and  P  is  the  middle  point  of  the  chord 
AB'.  In  a  similar  manner,  he  sees  that  the 
middle  point  of  the  chord  AC,  tangent  to  M 
at  A,  is  also  a  point  in  the  locus. 

Again,  he  observes  that  as  B  moves  towards  M,  and  C  towards  N,  P  moves  to- 
wards A,  and  when  AC  =  AB,  P  is  at  A.  It  now  appears  probable  that  the  locus 
of  P  is  a  circumference.  Proceeding  on  this  hj-pothesis,  he  reasons,  that,  if  this  is 
true,  AP'  and  AP"  are  chords  of  the  locus,  and,  bisecting  them  with  perpendicu- 
lars, he  will  have  the  centre  of  the  locus.  Locating  0  thus,  he  observes  tliat  it 
appears  to  be  in  the  line  joining  the  centres  0'  and  O",  and  about  midwaj'  be- 
tween them.  This  leads  him  to  see,  whether,  by  assuming  the  middle  point  of 
0  0  '  as  the  centre  of  a  circle,  and  OA  as  a  radius,  he  can  prove  that  any  such 
line  as  BC  drawn  through  A  is  bisected  by  this  circumference,  as  at  P.  This 
jic  can  readily  prove  by  means  of  the  perpendiculars  OD,  O'D',  and  O'D", 
which  bisect  the  chords  AP,  AB,  and  AC.  For,  since  these  perpendiculars  are 
parallel,  and  00  =:  00",  D'D  =  DD" ;  whence  DP  =  AD",  and,  adding  AP 
to  each,  DA  =  PD",  or  DB  =  PD".  Adding  to  BD',  DP,  and  D"C  (=  AD'  = 
DP),  there  results  BP  =  PC,  and  the  hypothesis  is  true. 

Bat,  in  (jii'ing  this  problem  as  a  recitation,  the  student  will  proceed  as  follows: 
Letting  M  and  N  be  the  two  fixed  circumferences,  intersecting  at  A,  join  their 
centres  0'  and  0",  and  bisect  00"  as  at  O.  With  O  as  a  centre  and  OA  as 
a  radius,  describe  a  circle.  Then  is  this  circumference  the  locus  required. 
For,  let  BC  be  any  secant  line  passing  through  A,  we  may  show  that  P  is  the 
middle  point  of  BC-  [Having  done  this,  as  above,  and  shown  that  any  point 
not  in  this  circumference  is  not  the  middle  of  the  secant  line  passing  through  A, 
his  solution  is  complete.] 


^■)  87  o.  I^roh. — If  the  line  AB  is  divided  at  Cjfind  the  locus  of  P,  so 
that  angle  apC  =  angle  BPC 

S^^g's. — In  seeking  for  t1i£  solution,  the  following  would  be  a  natural  process. 
Drawing  any  line,  as  AB,  in  the  lower  part  of  the  figure,  taking  C,  any  point  in 
it.  and  conceiving  BP  and  AP  drawn  so  as  to  md^ie  equal  angles  vj'Wh  PC,  we 
would  naturally  discover  that,  if  a  circle  were  circumscribed  about  BPA,  PC 
produced  would  bisect  the  arc  below  AB.  Thus  we  discover  a  ready  method 
of  locating  P ;  i.  e.,  in  the  main  figure,  bisect  AB  by  a  perpendicular,  as  ED,  and 


*  Equal  circles  would  probably  have  sj)€Cial  relations. 


OF  LOCI. 


295 


with  any  point  on  ED  as  a  centre,  pass  a  circumference  through  A  and  B.  Through 
D  and  C  draw  a  line,  and  P  is  a  point  in  the  locus  (?).  Any  number  of  poinds 
can  be  found  in  this  way ;  and, 

having  found  a  few,  as  P,  P',  P",  | 

P'",  etc.,  the  situation  of  these 
will  suggest  that,  probably,  the 
locus  is  the  circumference  of  a 
circle  whose  centre  is  in  A B  pro- 
duced. If  this  should  be  the  fact, 
CP  is  a  chord  of  that  circle,  and, 
erecting  a  perpendicular  at  the 
middle  point  of  CP,  its  inter- 
section with  AB  produced,  as 
O,  will  be  the  centre  of  the 
locus  (?).  "We  will  now  endeavor 
to  prove  that  any  point  in  this 
circumference,  as  P',  is  so  situated 
that  BP'C  =  CP'A,  and  that  no 
point  out  of  this  circumfer- 
ence has  this  property.     We  can 

readily  show  that  the  angle  OPB  -  OCR  —  BPC  =  OCR  —  CRA  (?).  But 
PAC  =  OCR  -  CRA  (?).  .  • .  Triangle  ORB  is  similar  to  OPA  (?),  and 
OA  ■:  OR  :  :  OR  :  OB,  or  OA  :  OC  : :  OC  :  OB.  Now,  for  any^omX  in  this  cir- 
cumference, as  R',  we  shall  have  OA  :  OR'  :  :  OR'  :  OB,  since  OR'  =  OC,  and  OA, 
OC,  and  OB  are  constant.  Hence,  wherever  R'  is  taken  (in  this  circumference), 
the  triangle  ORB  is  similar  to  ORA,  angle  OR'B  =  P'AB,  and  BR'C  =  CP'A. 
Finally,  that  no  point  out  of  this  circumference  possesses  this  property  is  evi- 
dent, since  the  distance  of  such  a  point  from  O  would  not  equal  OC,  and  the 
angle  OPiB  (Pj  being  such  a  point)  would  not  equal  PiAB. 


SKG.  Prob. — In  a  fixed  circle^  any  two  chords  intersect  at  right 
-^   angles  in  a  fixed  point ;  find  the  locus  of  the  centre  of  the  chord  join- 
^ying  their  extremities.     Give  the  proof. 

877.  Prob. — Fi7id  the  locus  of  the  point  in  space  equidistant 
from  three  given  p>oints.     Give  the  proof. 

878.  Prob. — Find  the  locus  of  the  point  in  space  equidititant 
from  tico  given  points.     Give  the  proof. 


879.  Prob. — Find  the  locus  of  the  point  in  a  plane  such  that 
the  difference  of  the  squares  of  the  distances  from  it  to  two  fixed 
points  tuithout  the  plane  shall  he  constant. 

Sug's. — Conceive  the  two  points  without  the  plane  joined  by  a  right  line, 
and  a  perpendicular  to  this  line  drawn  from  either  extremity  of  it ;  the  point 
where  this  perpendicular  pierces  the  plane  is  a  point  in  the  locus  (?).  The  re- 
quired locu*  is  two  parallel  straight  lines. 


296  INTRODUCTION  TO  MODERN  GEOMETRY. 

880.  Proh. — Find  the  locus  of  the  middle  point  of  a  straight 
line  of  constant  length,  whose  extremities  remain  in  two  lines  at 
right  angles  to  each  other,  lut  which  are  not  in  the  same  plane.    Give 

'^  \  the  proof. 

881.  I*roh. — Find  the  locus  of  the  point  equidistant  ffom  two 
fixed  p  la  n  es.    Give  proof. 

Sug's.— Consider,  1st,  When  the  fixed  planes  are  parallel ;  and  2d,  when  they 
intersect. 


882.  JProh- — What  locus  is  the  intersection  of  a  plane  and  the 
surface  of  a  sjihere  9    Give  proof. 


883.  I^rob. —  Wliat  locus  is  the  intersection  of  the  surfaces  of 
tico  given  spheres  9 

884.  JProb. — Find  the  locus  of  the  point  in  space  such  that  the 
ratio  of  its  distance  from  a  given  right  line  to  its  distance  from  a 
fixed  point  in  that  line  is  constant. 


SECTION  IL 

OF  SYMMETRY. 

885.  Def. — Two  points  are  said  to  be  symmetrical  with  resjoect  to 
a  third,  when  the  right  line  joining  the  two  points  is  bisected  by  the 
point  of  reference,  called  the  Centre  of  Symmetry. 

886.  Def. — Two  loci,  or  two  parts  of  the  same  locus,  are  sym- 
p,  ^^  metrical  luith  respect  to    a  p>oint,   when 

/     \/^"^~^^         every  point  in  one  has  its  symmetrical  point 
■'^        'C^^^^^^^:—^       ^^^  ^^  other. 

Xa)             b^C^^.^-'^^B^  Ill's. — In  (a)  P'  is  symmetrical  with  P  in  re- 

m'  spect  to  S,  if  S  is  the  middle  point  of  PP'.    In  (&) 

C*)  we  observe  that  the  semi-circumference  A^ti'B  is 

^                                '  symmetrical  with  the  semi-circumference  A7/iB, 

p/^]\,.^^     ^-<<I^^p'  ^^  respect  to  the  centre  S  ;  for  any  point  P  in  the 

p/„.__--;;;:^^^^^^^---^  #  latter  has  a  symmetrical  point  P'  in  the  former. 

g/-^^^'^  ^^^  ^^^^  -^^  ^^)  ^^^  triangle  A'SB'  is  symmetrical  with  ASB 

Pie.  455.  "1  respect  to  S  (?). 


OF   SYMMETRY. 


297 


887.  TJieo. — The  symmetrical  of  a 
straight  line,  ivith  resjMct  toa  jpoint,  is  an 
equal  straight  line. 

Sug's. — We  commence  by  assuming  AS  = 
AS,  and  B'S  =  BS,  and  drawing  B'A'.  We  then 
have  to  show  that  any  point  in  AB,  as  P,  has  its 
symmetrical  point  P'  in  B'A'. 


Fig.  456. 


888.  Theo. — The  symmetrical  of  an  angle,  luith 
2)oint,  is  an  equal  angle. 

Sug's. — To  show  the  symmetrical  of  AOB,  with  respect  to 
S,  take  AS  =  AS,  B'S  =  BS,  OS  =  OS,  and  draw  O'B',  O'A'. 
Then  show  that  any  point  in  OA  has  its  symmetrical  in  O'A', 
and  any  point  in  OB  has  its  symmeti'ical  in  OB'.  Hence, 
A'O'B'  is  the  symmetrical  of  AOB,  with  respect  to  S. 

Then  apply  AOB  to  A'O'B'  and  show  that  these  symmetricals 
are  equal. 


Fig.  457. 


889.  JProh. — Having  given  a  polijgon,  to 
draiv  its  symmetrical  ivith  respect  to  a  given 
point. 


890'  Tlieo. — Any  polygon  is  equal  to  its 
symmetrical  with  respect  to  a  given  point.  Fig.  458. 

Sug's. — Proof  by  revolving  the  figure  about  S,  keeping  it  in  its  own  plane, 
each  line,  as  AS,  ES,  et€.,  passing  through  180°. 

891.  Def. — The  lines  AS,  ES,  BS,  etc.,  are  radii  of  symmetry. 

892.  Def. — Two  points  are  symmetrical  with  respect  to  a  straight 
line  in  the  same  plane,  when  the  straight  line  which  joins  them  is 
bisected  at  right  angles  by  the  line  of  reference,  called  the  Axis  of 
Symmetry. 

893.  Def. — Two  loci,  or  two  parts  of  the  same  locns,  are  sym- 
pietrical  loith  resjyect  to  a  straight  line  when  every  point  in  the  one 
has  its  symmetrical  point  in  the  other. 

Ill's.— Thus  in  {a)  P  and 
P'  are  symmetrical  points  i'^ 

with  respect   to  the  riglit  *  I 

line  X'X,  P's  =r  Ps  and  X'X \ 

X  *'■  X 

is  perpendicular  to  PP'.   So  j 

the  part  of  (&)  above  X'X  is  I 

symmetrical  with  the  part         («)     -?' 
below,  i.  e.,  the  curve  is  sym- 
metrical with  respect  toX'X.  Fig.  459. 


298 


INTEODUCTION  TO  MODERN  GEOMETRY. 


894.  Theo. — The  symmetrical  of  a  straight  line,  with  respect  to 
a  rectilinear  axis  of  symmetry,  is  an  equal  right  line. 


P^B 


Dem. — Let  AB  be  any  straight  line,  and  X'X  the 
axis.  Let  fall  the  perpendiculars  A«  and  B^,  and 
produce  them  till  k's  =  As,  and  B's  =  Bs.  Then  is 
AB  =  AB',  the  symmetrical  of  AB.  For,  taking  P, 
(Oil/  point  in  AB,  letting  fall  Ps,  and  producing  it  to 
P',  the  point  P'  is  symmetrical  with  P  ;  since  re- 
volving sA'B's  upon  X'X,  AB'  will  coincide  with 
AB,  and  P'  will  fall  at  P.  Hence  A'B'  =  AB,  and 
every  point  in  AB  has  its  symmetrical  point  in 
A'B'  {893). 

895.  Cor.  1. — If  two  straight  lines  intersect,  their  symmetricals 
intersect,  and  the  points  of  intersection  are  symmetrical. 

The  student  should  show  how  this  follows  from  the  proposition. 

896.  CoR.  2. — Two  rectilinear  symmetricals  meet  the  axis  in  the 
same  jjoi7it,  and  malce  cqiial  angles  therewith. 

Student  give  proof. 

89'^.  Def. — A  Trapezoid  like  abb'a',  haying  its  non-parallel  sides 
equal,  is  called  Isosceles. 


898.  Theo. — The  symmetrical  of  an  angle, 
loith  respect  to  a  rectilinear  axis  of  symmetry, 
is  an  equal  angle. 

Sug's. — The  student  should  be  able  to  give  the  dem- 
onstration from  the  figure,  in  a  manner  altogether 
similar  to  the  preceding  ;  or,  drawing  AB  and  A'B',  he 
can  base  it  upon  the  preceding. 


^,9.9.  JPr oh.— Having  given  a  poly- 
gon, to  draw  its  symmetrical  with  respect 
to  a  given  axis. 


Fig.  462. 


900.    Theo. — Any  plane  figure  is 
^      equal  to  its  symmetrical,  with  reference 
to  a  rectilinear  axis. 


Proof  by  applying  one  to  the  other  by  revo- 
lution. 


& 


OF  SYMMETRY.  299 

901k  Def. — Two  loci  in  space,  or  two  parts  of  the  same  locus 
(planes  or  solids),  are  symmetrical  with  respect  to  a  point,  when  every 
point  in  one  has  a  corresponding  point  in  the  other,  such  that  the  line 
joining  them  is  bisected  by  the  point  called  the  centre  of  symmetry. 

Ill's. — Symmetrical  triednils  (Part  II.,  432)  afford  an  illnstratiou  of  solids 
symmetrical  with  respect  to  a  point — the  vertex.  The  two  hemispheres  into 
wliicli  a  great  circle  divides  a  sphere  are  symmeti'ical  parts  of  the  solid  (sphere) 
with  reference  to  the  centre. 

002.  Def. — Two  points  in  space  are  symmetrical  with  respect  to 
a  plane  called  the  Plane  of  Symmetry,  when  the  line  joining  the 
points  is  perpendicular  to  the  plane  and  bisected  by  it. 

903.  Def. — Two  loci  in  space  (planes  or  solids),  or  tw^o  parts  of 
the  same  locus,  are  symmetrical  with  respect  to  a  plane  when  every 
point  in  one  has  its  symmetrical  point  in  the  other. 

904:.  Def. — The  corresponding  (symmetrical)  parts  of  symmet- 
rical figures  are  called  Homologous  parts. 


905.  Theo. — The  symmetrical  of  a  right  line,  with  res2yect  to  a 
plane,  is  an  equal  right  line. 

Dem. — Let  A B  be  any  right  line,  and  MN  the  plane 

of  symmetry.    Let  fall  the  perpendiculars  B<^,  Aa,  upon  a, 

the  plane,  produce  tliem,  making  B'b  =   Sb,  IK'a  =  Aa,  ^^^-^ 

and  join  A'  and  B'.    Then  A'B'  =  AB,  and  is  its  sym-  i^ 

metrical.    For  ABB 'A'  being  a  plane  figure  (?)  and  ah,  l*f — i \ — \ -7 

the  intersection  of  this  plane  witli  MN,  being  a  right  /      \      !     ;^       / 

line  bisecting  AA'  and  BB'  at  right  angles  (?),  we  may  /             ''|     j      / 

revolve  a^B'A'  upon  ah  and  bring  A'B'  into  coincidence     ^ \ \ — | — ^ 

with  AB.      Hence  A'B'  —  AB.    Again,  P  being  any  \      \)^ 

point  in  AB,  draw  PP'  perpendicular  to  ah,  and  upon  U^ 

revolution  P' will  fall  in  P,and  P's  =  Ps.  Hence,  every  a' 

point  in  AB  has  its  symmetrical  in  A'B',  and  the  latter  ^m.  4G3. 
line  is  the  symmetrical  of  the  former. 

906.  Cor.  1. — A  right  line  and  its  symmetrical,  with  respect  to  a 
'plane,  ijierce  the  plane  at  the  same  point. 

Student  give  proof 


9011.  Theo. — Tlie  symmetrical  of  a  plane  angle,  with  resp>ect  to  a 

plane,  is  an  equal p>lane  angle. 


300 


INTRODUCTION  TO  MODERN  GEOMETRY. 


Dem. — Let  AOB  be  any  plane  angle,  and  MN  the 
plane  of  symmetiy.  Let  P  be  any  point  in  AO,  and 
Pi  any  point  in  OB.  Let  0'  be  the  symmetrical  of  O, 
P'  of  P,  and  Pi'  of  Pi  ;  then  is  AG'  the  symmetrical  of 
AO,  O'B'  of  OB,  and  angle  A'O'B'  of  AOB.  Now  by 
the  preceding  proposition  the  two  triangles  POPi,and 
P'O'Pi',  are  mutually  equilateral,  whence  AOB  =  its 
symmetrical  A'O'B'. 


Query. — When  will  the   triangle  'pop'    exist, 
when  not  ? 


and 


008.  TJieo. — Any  lilane  polygon  has  for  its  symmetrical,  loith 
reference  to  a  iMne,  an  equal  lylane  lyolygon. 


Sug's.— ABODE  being  any  plane  polj'gon,  and 
MN  the  plane  of  symmetry,  by  constructing 
A',B',C',D',E'  sj^mmetrical  with  A,  B,  C,  D,  E,  we 
have  by  the  preceding  propositions  A'B'C'EyE' 
equilateral  and  equiangular  with  ABODE  ;  whence 
it  only  remains  to  show  that  A'B'C'D'E'  is  a  plane 
(not  a  warped)  surface.  Let  F  be  any  point  in  the 
angle  AED,  draw  HI,  and  let  H'  and  I'  be  the  sym- 
metricals  of  H  and  I  {895).  Draw  H'l'.  Then  is 
the  symmetrical  of  F  in  H'l'  (?),  as  at  F'.  Now, 
every  point  in  HF  within  the  angle  BAE  has  its 
symmetrical  in  H'F'  {^05\  Thus,  by  taking  three 
points,  not  in  a  straight  line,  in  the  angle  BAE,  we 
can  show  that  their  syrametricals  are  in  the  plane 
B'A'E',  and  also  in  A'E'D'.  In  like  manner,  all 
the  angles  of  A'B'C'D'E'  can  be  shown  to  be  in  the  same  plane. 

000.  Cor. — If  two  x>lcines  intersect,  their  symmetricals  intersect, 
and  the  two  intersections  are  symmetrical  right  lines. 

The  student  should  show  how  this  grows  out  of  the  proposition. 


010.  TJieo. — The  symmetrical  of  a  diedral  is 
an  equal  diedral, 

Dem.  AOB  being  the  measure  of  the  diedral  A-OC-B, 
and  A'-O'C'-B'  the  symmetrical  diedral,  and  0'  the  sym- 
metrical of  O,  the  symmetrical  of  AO  being  A'O',  the  angle 
A'O'C  is  right,  and  in  like  manner  B'O'  being  the  symmet- 
rical of  BO,  B'O'C  is  right.  But  BOA  =  B'O'A'  (?),  whence 
the  diedrals  are  equal. 


Fig.  466. 


OF   SYMMETRY. 


301 


Oil.  HieO' — Two  'pohjedrons,  symmetrical 
with  respect  to  a  plane,  have  their  faces  equal, 
each  to  each,  and  their  homologous  solid  angles 
symmetrical. 

Sug's. — This  is  an  immediate  consequence  of  preced- 
infi^  propositions.  Thus  E'  being  the  symmetrical  solid 
liomologous  with  E,the  homologous  plane  foces  includ- 
ing them  are  equal  {908).  Again,  the  facial  angles 
being  equal,  but  not  similarly  disposed,  the  solid  angles 
are  symmetrical. 


,^ 

//' 

0 

ri 

}A^ 

__  ^ 

M      i      i 

:B    ; 

/n 

11/ 

'H 

■i-irf^ 

/f 

\~\ 

'  l- 

H' 

V 

Fig.  467. 


912.  Cor. — Two  symmetrical  2)olyedrons  can 
he  decomjjosed  into  the  same  nnmher  of  tetraedrons,  symmetrical  each 
to  each. 

For  we  can  decompose  one  of  the  polyedrons  into  tetraedrons  having  for 
their  common  vertex  one  of  the  vertices  of  this  ]3olyedron,  and  each  of  these 
tetraedrons  will  have  its  symmetrical  in  the  other. 


913'  Theo. — Two  symmetr iced  polyedrons  are  equivalent. 

Dem.— From  the  last  corollary  it  will  appear  that  it  is  sufficient  for  the  de- 
monstration of  this  proposition  to  show  that  two  symmetrical  tetraedrons  are 
equivalent  (?).  Let  S-ABC,  and  S-ABC  be  two  tetraedrons  symmetrical 
with  respect  to  their  common  base.  They  have  a  common  base  and  equal  alti- 
tudes (?),  hence  they  are  equivalent. 


914:.  General  Scholium.— We  may  speak  of  two  loci,  or  two  parts  of  the 
same  locus,  as  symmetrical  with  respect  to  a  line  or  plane,  whenever  all  the 
points  in  one  have  symmetrical  points  in  the 
other,  even  though  the  line  joining  the  symmet- 
rical points  be  not  'perpendicular  to  the  axis,  or 
the  plane,  of  symmetrj"- ;  observing,  however, 
that  this  line  is  always  bisected  by  the  axis  or 
plane.  Thus,  tlie  ellipse  in  the  figure  is  symmet- 
rically divided  by  the  line  X'X,  since  every  point 
in  one  portion  has  a  symmetrical  point  in  the 
other,  as  Ps  =  P'«,  for  every  point  in  the  curve. 
In  such  a  case  the  parts  cannot  be  brought  into 
coincidence  by  simple  revolution :  one  part  must  be  reversed. 


Fig.  468. 


302 


INTRODUCTION   TO   MODERN   GEOMETRY. 


SECTION  III 

OF  MAXIMA  A>D  MINDIA. 

915.  Def. — A  Maxiniinn  value  of  a  magnitude  conceived  to 
vary  continuously  in  some  specified  way,  is  a  value 
ffi^^  "which  is  greater  than  the  preceding  and  succeeding 

values  of  the  magnitude. 


Ill's. — Thus,  suppose  in  a  given  circle,  a  chord  passing 
through  a  fixed  point,  P,  revolves  so  as  to  take  successively 
the  positions  1«,  26,  Ac,  3^, 4«, etc.  It  is  at  a  maximum  when 
it  passes  through  the  centre,  as  Ac.  The  chord  is  the  magni- 
tude which  is  conceived  to  vary  in  the  way  specified,  and  Ac 
is  a  value  greater  than  the  preceding  and  the  succeeding 
values.  Again,  conceive  a  circle  to  be  compressed  or  ex- 
tended, as  in  the  direction  mn,  so  as  to  take  the  forms  in- 
dicated by  the  dotted  lines,  its  area  will  be  diminished, 
the  perimeter  remaining  the  same.  That  is,  of  all 
figures  of  a  given  perimeter,  the  circle  has  the  maxi- 
mum area. 


Fig.  470. 


910'  Def. — A  Mininiuni  value  of  a  magnitude  conceived  to 

vary  continuously  in  some  specified  way,  is  a  value 
which  is  less  than  the  preceding  and  succeeding 
values  of  the  magnitude. 

Ill's. — Thus,  conceive  the  varying  magnitude  to  be  a 
straight  line  from  the  fixed  jwint  P  to  the  fixed  line  X'X  ; 
that  is,  suppose  such  a  line  to  start  from  some  position  PI. 
and  move  through  the  successive  positions  P2,  PA,  P3,  P4. 
PA  is  a  minimara,  since  it  is  less  than  the  preceding  and  succeeding  values. 


PROPOSITIOXS  COXCERXIXG  MAXDIA  AM)  MIXDIA. 

917*  Axiom. — Tlie  minimum  distance  between  tico poi?its  is  a 
straight  line. 

918.  TJieo. — The  minimum  distance  from  a  point  to  a  line  is  a 
straight  line  2)erpendicular  to  the  given  line. 

Student  give  proof. 

919.  Tlieo. — Tiie  maximum  line  which  can  he  inscribed  in  a 
given  circle  is  a  diameter. 


OF  MAXIMA  AND   MINIMA. 


303 


Proof  based  on  the  fact  that  the  hypotenuse  of  a  right  angled  triangle  is  the 
greatest  side. 


920.  Theo^ — The  sum  of  the  distances  from 
tiuo  points  on  the  same  side  of  a  line,  to  a  poi?it 
in  the  line,  all  heing  in  the  same  plane,  is  a  mini- 
mum lohen  the  lines  measuring  the  distances 
make  equal  angles  with  the  given  line. 

Student  prove  AP  +  BP  <  AP'  +  BP'. 


Fig.  472. 


921.  TJieo. — If  a  triangle  have  a  constant 
base  and  altitude,  its  vertical  angle  is  a  maxi- 
7num  iuhen  the  triangle  is  isosceles. 

SXFG. — By  what  is  the  vertical  angle  measured  ? 


Fig.  473. 


922.  TJieo. — The  lase  and  area  of  a  triangle  being  constant,  its 
perimeter  is  a  minimum  johcn  the  triangle  is  isos- 
celes. A 


Sug's. — The  area  and  base  being  constant,  the  vertex 
remains  in  a  line  parallel  to  the  base,  for  all  values  of  the 
other  sides.  The  figure  will  suggest  the  demonstration, 
which  is  based  on  the  fact  that  any  side  of  a  triangle  is 
less  than  the  sum  of  the  other  two. 


Fig.  414, 


923.  TTieo. — The    difference  between  the  distances  from  tico 
points  on  opposite  sides  of  a  fixed  line  toa  p)oint 
in  that  line,  is  a  maximum,  when  the  lines 
7neasuring  these  distances  make  equal  angles 
ivith  the  fixed  line. 

Sug's.     P'O  =  AP  -  AP';  but  PO  >  A'O  (=  A'P) 
—  A'P'. 


Fig.  475. 


Query.— Having  the  points  P,  P',  and  the  fixed  line 
given,  how  is  the  point  A  found  by  geometrical  construction  ? 


924.  Theo. — The  lengths  of  two  sides  of  a 
triangle  being  constant,  the  area  is  a  maximum 
lohen  the  iiicluded  angle  is  right. 


Fio.    476. 


304 


INTRODUCTION  t6   MODERN  GEOMETRY. 


925.  Tlieo. — The  sum  of  tico  adjacent  sides  of 
a  redaitgle  being  constant  (AB),  the  area  is  a  maxi- 
mum when  the  sides  are  eqiial. 


ISOPEREttETRY. 

026'  Isoperinietric  Figures  are  such  as  have  equal  perim- 
eters, i.  e.,  bounding  lines  of  equal  length. 

Problems  in  isoperimetry  are  a  species  of  problems  in  Maxima  and 
Minima.  Thus,  of  all  figures  whose  perimeters  are  m  (say  10 
inches),  to  find  that  which  has  the  greatest  area,  is  a  problem  in 
isoperimetry.  Again,  what  must  be  the  form  of  a  pentagon  whose 
perimeter  is  m,  in  order  that  its  area  may  be  a  maximum  ? 

027'  Theo. — Of  isoperimetric  triangles  with  a  constant  base,  the 
isosceles  is  a  maximum. 

Sug's.— By  means  of  the  figure  to  Theorem  {921),  we  can  readily  show 
that  any  triangle  having  the  same  base  as  the  isosceles  triangle,  and  its  vertex 
either  in  or  beyond  the  line  through  the  vertex  of  the  isosceles  triangle  and  par- 
allel to  its  base,  has  a  greater  perimeter  than  the  isosceles  triangle.  Hence,  the 
isoperimelric  triangle  on  the  given  base  has  its  vertex  below  this  parallel,  ex- 
cept when  isosceles  ;  and  consequently  the  isosceles  is  the  maximum. 

028.  Cor. — Of  isoj^erimetric  triangles,  tlie  equilateral  has  the 
maximum,  area  (?) 


020.  JProb. — Given  any  triangle  with  a  constant  lase,  to  con- 
struct the  maximum  isoperimetric  triangle. 


030.  ProT). — Given   any  triangle,  to   construct  the  maximum 
isoperimetric  triangle. 

031.  Theo. — Of  isoperimetric  quadrilaterals, 
the  square  has  the  maxinmm  area. 

Dem.— Let  ABCD  be  any  quadrilateral.  If  AD  is  not 
equal  to  DC,  ADC  can  be  replaced  by  the  isosceles 
isoperimetric  triangle  AD'C,  and  the  area  of  the  quadri- 
lateral increased.  So  ABC  can  be  replaced  by  AB'C.  There- 
fore AB'C'D  >  ABCD.  In  like  manner  if  AD'  is  not  equal 
to  AB'  ,  D'AB'  can  be  replaced  by  the  maximum  isoperi- 
metric triangle  DAB'.  So  also  D'CB'  can  be  replaced  by 
D'C'B'.  Therefore  AB'C'D'  >  AB'CD'  >  ABCD.  Now, 
A'B'C'D'  is  a  rhombus  (?),  and  the  student  can  show  that 
the  square  on  A'B'  is  greater  than  any  rhombus  with  the 
same  side. 


Fig.  478. 


ISOPERIMETRY. 


305 


932.  JProb. — Having  given  a  quadrilateral,   to    constrtict  the 
maxivium  isoijerimetric  quadrilateral. 


933.  Theo. —  Of  isoperir}ietric  quadrilaterals  luitli  a  constant 
base,  the  maximum  has  its  three  remaining  sides  equal  each  to  each, 
and  the  angles  ivhich  they  include  equal. 

Dem. — Let  ABCD  be  the  maximum  isoperimetric  quadrilateral  on  the  base 
AD,  then  AB  =  BC  =  CD,  and  angle  ABC  =  BCD.  For,  if  AB  is  not  equal  to 
BC,  draw  AC,  and  replacing  the  triangle  ABC  with 
its  isoperimetric  isosceles  triangle,  we  shall  have  a 
quadrilateral  isoperimetric  with  ABCD,  and  greater 
than  ABCD,  i.  e.,  greater  than  the  maximum,  which 
is  absurd. 

Again,  if  angle  ABC  is  not  equal  to  BCD,  let  ABC 
<  BCD,  whence  BCE  <  EBC,  and  BE  <  EC.  Take 
EF  =  EC,  and  EG  =  EB,  whence  the  triaugles  FEG 
and  BEC  are  equal,  and  FC  =  BC.  Also,  since  AB 
+  BC  +  CD  =  AE  +  ED  -  (EB  +  EC)  +  BC,  and 

AF  +  FG  +  CD  -  AE  +  ED  -  (FE  +  EC)  +  FG,  it  follows  that  AFGD  and  ABCD 
are  isoperimetrical,  and,  since  ABCD  =  AED  -  BEC,  and  AFGD  =  AED  -  FEG, 
that  AFGD  and  ABCD  are  equal.  Therefore,  AFGD  is  a  maximum,  and  by  tlie 
preceding  part  of  the  demonstration  AF  =  FG  =  BC  =  AB,  which  is  absurd; 
and  there  can  be  no  inequalit}'-  between  angles  ABC  and  BCD. 


Fig.  479. 


934.  Theo. — Of  isoperimetric  polygons  of  a  given  number  of 
sides,  the  regular  polygon  has  the  maximum  area. 

Dem. — First,  the  polygon  must  be  equilateral ;  for, 
if  any  two  adjacent  sides,  as  AB,  BC,  are  unequal,  the 
triangle  ABC  can  be  replaced  by  its  isoperimetric 
isosceles  triangle,  and  thus  the  area  of  the  polygon 
be  increased. 

Second,  the  pol3^gon  must  be  equiangular;  for,  if 
any  two  adjacent  angles,  as  B  and  C,  are  unequal,  the 
quadrilateral  ABCD  can  be  replaced  by  its  isoperime- 
tric quadrilateral  with  B  =  C,  and  thus  the  area  of  the  polygon  be  increased. 


Fig.  480. 


^  93 S.  TJieo. — Of  isoperimetric  regular  polygons,  the  one  of  the 
greater  number  of  sides  is  the  greater. 

Dem. — Let  ABC  be  an  equilateral  (regular)  triangle. 
Join  any  vertex,  as  A,  with  any  point,  as  D,  in  the  opposite 
side.  Replace  the  triangle  ACD  with  the  isosceles  isoperi- 
metric triangle  AED.  Then  is  the  quadrilateral  ABDE  > 
the  triangle  ABC. 

But,  of  isoperimetric  quadrilaterals,  the  regular  (the 
square)  is  the  greater.  Hence,  the  regular  quadrilateral  (the 
square)  isoperimetric  with  the  trianulc  ABC,  is  greater  than 

^20 


306 


IXTEODUCTION   TO   MODERN   GEOMETRY, 


the  triangle.    In  the  same  manner  the  regular  pentagon  isoperimetric  with  the 
square  can  be  shown  greater  than  the  square ;  and  tJius  on,  ad  libitum. 

936.  Cor. — Of  plane  isojjerimetric  figures,  the  circle  has  the 
maximum  area,  since  it  is  the  limiting  form  of  the  regular  polygon, 
as  the  number  of  its  sides  is  indefinitely  increased. 


.:,v^ 


SECTION   IV, 

OF  TRAXSYERSALS. 


^R 


f>,?7.  Def. — A  Transversal  is  a  line  cutting  a  system  of  lines. 
A  transversal  of  a  triangle  is  a  line  cutting  its  sides  ;*  it  either  cuts 
two  sides  and  the  third  side  produced,  or  the  three 
sides  produced.     In  speaking  of  the  transversal 
of  a  triangle  (or  polygon),  the  distances  on  any 
side  (or  side  produced)  from  the  intersection  of 
the  transversal  with  that  side  to  the  angles,  are 
Segments.     Of  these  there  are  six.    Adja- 
cent segments  are  such  as  have  an   ex- 
tremity of  each  at  the  same  point.     Xon- 
adjacent  segments  are  such   as  have  no 
extremity  common. 

Ii.l's.    TR  is  a  ti-ansversal  of  the  triangle 
P,g  4^  ABC  ;  rtA,  aC,  bC,  bS,  ^A,  cB  are  adjacent  seg- 

ments two  and  two  ;  aC,  6B,  cA,  and  aA,  bC,  cB 
are  the  two  groups  of  non-adjacent  segments. 

938.  The  two  Fukdamextal  Propositions  of  the  Theory 
OF  Transversals. 

939.  Theo. — Tlie  product  of  three  non-adjacent  segments  of  the 
sides  of  a  triangle  cut  hy  a  transversal,  is  equal  to  the  j^roduct  of 
the  other  three. 

Dem. — ABC  being  cut  by  the  transversal  TR,  aA  x  6C  x  cB  =  aC  x  Z/B  x  ck. 
Draw  BD  parallel  to^AC,  and  from  the  similar  triangles  we  have 

DB      aC 


rtA      cA 

Bb=^'''^"^^7;B 


bO 


whence,  multiplying, 
rtA  _  rtC  X  cA 
hB  ~  bt   x   cB' 

or  aA    X    6C    x    cB 

«C  X  6B  X  cA. 


Fio.  483. 


*  Or.  sides  produced— this  expression  bein?  usually  omitted  in  higher  Geometry 
are  to  be  considered  indefinite  unless  limited  in  the  problem. 


all  lines 


OF   TRANSVERSALS.  307 

940.  Cor. — Conyerselj,  If  tliree i^oints  le  tahen  in  the  sides  of 
a  triangle  (as  a,  b,  c)  such  that  the  product  of  three  non-adjacent  seg- 
ments equals  the  product  of  the  other  three,  the  points  are  in  the 
same  straight  line. 

For,  passing  a  line  through  a  and  h,  let  it  cut  the  third  side  in  c  .  Then,  by 
the   proposition,  aA  x    hC    x    c'B  =  aC    x    bB    x    c'A.      But,  by  hypothesis 

cB       cA 

aA  X  bC  X  cB  =  aC  X  bB  X  cA.  Whence  -—-  =  -—-,  and  c  and  c'  must  coincide. 

cB      c  A ' 

ScH. — This  theorem  is  known  among  mathematicians  as  77ie  Ptolemaic  TJieo- 

rem,  and  is  usually  attributed  to  Claudius  Ptolemy,  an  Egyptian  mathematician 

and  philosopher  who  flourished  in  Alexandria  during  the  first  half  of  the  second 

century.    But  it  is  thought  to  be  more  properly  due  to  Menelaus,  who  lived  a 

century  before  Ptolemy. 


041.  TJieo. — The  three  angle-transversals*  of  a  triangle,  2)assiug 
through  a  common  point,  divide  the  sides  into  seg- 
7ne7its  such  that  the  product  of  three  non-adjacent 
segments  equals  the  product  of  the  other  three.  ^/Ln\ 


^-?-\ 


Dem. — From  the  triangle  ACc  cut  by  the  transversal  r«B, 
we  have  aA  x  CO  x  cB  =:  AB  x  aC  x  Oc ;  and  from  CBc 
cut  by  *A,  Oc  X  iC  X  AB  —  CO  x  bB  x  cA.  Multiplying, 
we  obtain  aA  x  6C  x  c  B  =  aC   x  Z*B  x  cA. 

942.  Cor.  1. — Conversely,  If  the  three  angle- 
transversals  of  a  triangle  divide  the  sides  into  seg- 
ments such  that  the  product  of  three  non-adjacent      "      fig.  4^. 
segments  equcds  the  product  of  the  other  three,  the 
transversals  pass  through  a  com^non  p)oint. 

For,  the  sides  being  divided  at  «,  b,  and  c,  so  that  aA  x  &C  x  cB  =  ^tC  x  bB  x 
cA,  draw  Cc,  and  A&,  and  let  O  be  their  intersection.  Now,  let  a'  be  the  point 
in  which  BO  cuts  AC.  Then,  by  the  proposition,  a' A  x  Z'C  x  cB  ==  rt'C  x  //B  x  cA. 

Whence  -rr  =  -77^,  and  a  and  a'  coincide. 
a'A       a'Q 

'  943.  Cor.  2. — If  any  one  of  the  sides  is  bisected,  the  line  joining 
the  other  p>oints  of  division  is  parallel  to  this  side. 

For,  let  6C  =  bB.     Then  «A  x  6C  x  cB  =  aC  x  bB  x  cA,  becomes 
«A  X  cB  =  aO  X  cA  ;  or  Aa  :  «C  : :  Ac  :  cB. 

Query. — How  does  this  apply  to  the  second  figure  ? 

944.  Cor.  3. — If  the  line  joining  two  points  of  division  is  j^^r- 
allel  to  the  third  side,  the  latter  side  is  Usected. 

*  The  transversals  passing'  through  the  angles. 


308  INTRODUCTION   TO   MODERN   GEOMETRY. 

For,  if  ah  is  parallel  to  AB,  aC  :bC:  :ak  :  6B,  -whence  aC  x  JB  =  6C  x   aA. 
And,  since  rtA  x  6C  x  oB  —  aC  x6B  xcA,  cA  =  cB. 


.9d?J.  We  will  now  give  a  few  problems  to  illustrate  the  use  of 
the  theory  of  transversals. 

04:6.  J^rob. — To  show  that  the  medial  lines  of  a  triangle  j^ciss 
through  a  common  'point. 

Solution. — Since  aA  =  (tC,  bC  =  JB,  and  cB  =  cA,  by  mul- 
tiplying, we  have  a  A  x  bC  x  cB  =  aC  x  bB  x  cA  ;  -whence 
by  the  last  corollary  these  transversals  pass  through  a  com- 
mon point. 


047.  JProh. — To  shoio  that  the  bisectors  of  the  angles  of  a  tri- 
angle pass  through  a  common  point. 

Solution. — In  the  last  figure  let  «B,  5A,  cC  be  the  bisectors. 
^,         rtA       AB    bO      AC     cB       CB  ,  .  ,   .        «A  x  5C  x  cB 

T^^^   ^=CB'Z;B=AB'^=AC'  "^"It^Ply^^g'  aZxbBx  cA  =  ^'  ^^ 

rtA  x  &C  X  cB  =rtC  X  6B  x  cA.    Therefore  these  transversals  pass  through  a 

common  point. 


048.  JProb. — To  shoio  that  the  altitudes  of  a  triangle  yass 
through  a  common  point. 

Sug's. — In  the  last  figure,  if  rtB,  bk,  cC,  -were  the  perpendiculars,  there  would 
^     .    .,         .       ,        .  .      rtA        AO    JC       CO     cB       OB 
be  three  pairs  of  similar  triangles  givmg  ^  =|^'   M  =  AO'  ^  ==  CO  ' 

whence,  as  in  the  last.   / »  cc  A  X  t  L  X  c£  £,   vp  v  aJ^^x  ,7  0  , 

040.  Proh. — To  shoiu  that  the  angle-transversals  terminating  in 

the  points  of  tangcncy  of  the  sides  of  the  triangle  with  its  inscribed 
circle,  2)ciss  through  a  common  jwint. 

Sug's.— In  the  last  figure,  if  a,  b,  c  were  the  points  of  tangency  we  should 
have  rtA  =  cA,  bC  =  aC,  cB  =bB  ;  -whence  rtA  x  ^>C  x  cB  =  aC  x  JB  x  ck. 
AVhich  shows  that  the  transversals  pass  through  a  common  point. 

OoO.  Theo. — If  two  sides  of  a  triangle  are  divided  proportion- 
c  ally,  starting  from  the  vertex,  the  angle-trans- 

versals from  the  extremities  of  the  other  side 
to  the  corresponding  points  of  division,  in- 
tersect in  the  medial  line  to  this  third  side. 


Dem. — Since  AC  and  CB  are  divided  proportion- 
ally at  a  and  a\  rtA  x  a'C  =  aC  x  rt'B  ;  and  as 
DB  =  DA,  rtA  X  rt'C  X  DB  =  rtC  x  rt'B  x  DA, 
the   angle-transversals    Art',  Brt  intersect  in   CD. 

The  same  may  be  sho-wn  of  any  other  angle-transversals  from  A  and  B,  dividing 

CB  and  CA  proportionally. 


OF   TRANSVERSALS. 


309 


9S1'  Cor. — In  any  iraj^ezoid  the  transversal  j^^^ssing  through 
the  intersection  of  the  diagonals,  and  the  intersection  of  the  non- 
parallel  sides,  bisects  the  parallel  sides. 


SuG. — Joiniug  cut'  iu  the  last  figure,  CD  is  such  a  transversal.    The  student 
will  readily  see  the  connection  with  the  proposition. 


[%'^ 


OS 2.  JProb.— Through  a  given  point  to  draiu  a  line  which  shall 
meet  two  given  lines  at  their  intersection  in  an  invisible,  inaccessible 
2^0  int. 

Solution. — Let  Mm,  N?i  be  the  two 
given  lines  which  meet  in  the  invisible, 
inaccessible  point  S,  and  P  the  given 
point  through  which  a  line  is  to  be  lo- 
cated which  will  meet  Mw,  Hn  in  S. 
Through  P  draw  any  convenient  trans- 
versal, as  BF,  and  any  other  meeting  this, 
as  AF.  Now,  considering  MS  as  a  trans- 
versal of  the  triangle  CDF,  we  have 
AF  X  BC  X  SD  =  AD  X  BF  X  SC ;  whence 
SD  _  AD  X  BF 
SC 

HP  _SD_  AD 
PC 


AF  X  BC* 

parallel  to  BF,  we  have 


But,   HD  being  drawn 


Fio. 


BF 


or  HD  = 


AD 


BF 


PC 


SC       AF  X    BC  AF  X    BC 

whence  HD  is  known,  as  AD,  BF,  PC,  AF,  BC  can  be  measured.    Tlie  points  P 
and  H  determine  the  required  line. 

953.  Def. — The    Complete   Quaclrilateral  is  the  figure 
formed  by  four  lines  meeting  in  six 
points.     The   complete  quadrilateral 
has  three  diagonals. 

III. — ABCDEF  is  a  complete  quadrilat- 
eral, and  its  diagonals  are  CF,  BD,  and  AE, 
the  latter  being  spoken  of  as  the  third  or 
exterior  diagonal. 


Fig.  488. 


954.  Tlieo. — The  middle  points  of  the  three  diagonals  of  a  complete 
quadrilateral  are  in  the  same  straight  line. 

Dem.— m,  n,  o  being  the  centres  of  the  diagonals  of  the  complete  quadrilat- 
eral, in  the  preceding  figure,  are  in  the  same  straight  line.  Bisect  the  sides  of 
the  triangle  FDE,  as  at  I,  N,  L,  and  draw  IN,  IL,  LN.  Since  IN  is  parallel  to  BE, 
and  bisects  DF  and  DE,  it  also  bisects  DB  (?)  and  hence  passes  through  n.  For 
like  reasons  IL  passes  through  m,  and  LN  through  o.  Now,  AC  being  a  trans- 
versal of  the  triangle  FDE  gives  CD  x  BE  x  AF  =  CE  x  BF  x  AD.     Therefore, 


310  ES'TRODUCTION   TO   MODERN   GEOMETRY. 

noticing  that  iCD  =  ;/J,  ^BE  =  ;jN,  {AF  =  oL,  ^CE  =  mL,  ^BF  =  nl,  and 
MO  =  oN,  we  have  ml  x  71N  x  oL  =  mL  x  fi\  x  oN.  Hence  these  thiee  points 
m,n,o  lie  in  a  transversal  to  the  triangle  ILN. 


SECTION  V. 

HARMONIC  PROPORTION  AND  HARMONIC  PENCILS. 

^oo»  Def. — Three  quantities  are  in  Harmonic  Projyortion  when 
the  difference  between  the  first  and  second  is  to  the  difference  be- 
tween the  second  and  third,  as  the  first  is  to  the  third. 

III. — 6,  4,  3  are  in  harmonic  proportion,  since  6  —  4:4— 3::6:3.  In  gen- 
eral, a,b,  c  are  in  harmonic  proportion,  \i  a  —  b  :  h  —  c  : :  a  \  c. 


9o6.  Theo. — If  a  given  line  he  divided  infernally  and  externally 
in  the  same  geometric  ratio,  the  distance  beticeen  thepoitits  of  division 
is  a  harmonic  mean  between  the  distances  of  the  extremities  of  the 
given  line  from  the  point  not  included  between  them. 

Dem.  —  Let  AB  be  the  given  line ;    and  let  O  and  0'  be  so  taken  that 

AO  :  BO  :  :  AO'  :  BO' ;    then    is    00'    a   har- 

-— ~ — g ^7 —      monic  mean  between  AO'   and   BO'.     For 

AO  =  AO'-00',    and    BO  =  00'  -  BO' ; 
Fig-  490.  whcDce  AO',  00',  and    BO'  are  such   that 

AO'—  00' :  CO'  —BO'  :  :  AO' :  BO',  that  is,  tliey  are  in  harmonic  proportion. 


9J7.  Cor.  1. — AO,  AB,  and  AO'  are  in  harmonic  proportion^  i.e.; 
AB  is  a  harmonic  mean  between  AO  and  AO'. 

For  AB  -  AO  (=  BO) :  AO'  -  AB  (=  BO') :  :  AO  :  AO'. 

OoS.  Cor.  2. —  ]]lien  AOfOO',  BO'  are  in  harmonic  proportion, 
AO  X  BO'=r  BO  X  AO'. 

U^O.  Cor.  3. — Conversely,  }Ylien  a  line  is  divided  i?ito  three  parts 
K((ch  that  the  rectangle  of  the  extreme  parts  equals  the  rectangle  of  the 
mean  part  into  the  whole  line,  the  line  is  divided  harmonically. 

Thus,  let  AO' be  the  line,  and  AO  x  BO*  =  BO  <  AO';  then  AO:  BO  :  :A0'  :  BO', 
whence,  b}'  the  proposition,  00'  is  a  harmonic  mean  between  AO'  and  BO'. 

Def. — The  points  0  and  O'  are  called  Harmonic  Conjugates. 


HARMONIC    PROPORTION   AND    HARMONIC    PENCILS. 


311 


060.  Theo. — Iftivo  lines  M  clratvn,  one  hiseding  the  interior  and 
the  other  the  adjacent  exterior  angle 
of  a    triangle,  and   oneeting  the  op- 
posite  sidef^  they  divide  this  line  har- 
monically. 

SuG.— By  means  of  {358,  359,  Part 
II.)  the  student  will  be  enabled  to  establish 
the  relation  AO  :  BO  :  :  AO'  :  BO',  wlience, 
by  the  last  proposition,  AO',00',  BO'  are  in  harmonic  proportion. 

961'  Tlieo. — In    the    complete    quadrilateral,  any  diagonal   is 
Q  divided  harmonically  hy   the    other 

two, 

Dem. — Thus,  AFH  is  divided  harmoni- 
cally at  C  and  H.    For,  considering  BH  as 
'\.,  a  transversal  of  the  triangle  AGF,  we  have 

-■^H      HF  X  DC  X  BA  =  HA  X  DF  x  BC.      And 

CC,  AD,  FB  being  angle-transversals  of  the 
same  triangle,  we  have  CF  x  BA  x  DC  = 


Whence,  dividing,  ~  =  ^,  i.e.^  AH  is  divided  harmonically 
Gr       GA 


G 

FrG.  489. 

CA  x  BC  X  DF. 

Again,  if  CH  is  drawn,  CA,  CG,  CF,  CH  constitute  a  harmonic  pencil,  and  BH,  a 
transversal  of  it,  is  cut  harmonically  at  B,  I,  D,  H.  Finally,  if  F  and  I  be  joined, 
FH  (or  FA),  FB,  Fl,  FD  constitute  a  harmonic  pencil,  and  hence  CG  is  cut  har- 
monically at  C,  I,  E,  G. 

962.  Cor. — An  angle-transversal  of  a  triangle,  and  a  line  passing 
through  the  feet  of  the  other  uiigle-transversals,  divide  the  third  side 
harmonicall}-.  

963.  JProb.— Given  a  right  line  to  locate  ttvo  harmonic  coiijugate 
points. 

Solution.— Let  AB  be  the  line.  O  may  be  taken  at  pleasure  between  A  and 
B.  We  are  then  to  find  0',  so  that  AO  :  BO  : ;  AO' :  BO',  Taking  this  by  di- 
vision, we  have  AO  -  BO  :  BO  : :  AO'  -  BO'  (=  AB) :  BO'.  The  first  three  terms 
being  known,  the  other  can  be  constructed.  Or,  we  may  first  locate  0'  at 
pleasure,  and  then  find  O.  

964.  Theo. — If  from  the  given  point  C  in  a  line  the  distances 
CO,  CB,  CO'  he  taken  in  the  same  di- 
rection, so  that  CO  X  CO'  =  CB*  ?  and  if 
CA  =  CB  bo  taken  in  the  opposite  di- 
rection, AO'  will  he  divided  harmonically   at   O   and   B. 

Dem.  —  From    CO 
CB  -f-  CO(=AO):  CB 


O       B 
Fig.  492. 


X  CO'  =  CB^    we    readily    Avrite    CO 
-  CO  (=  BO) : :  CO'  +  CB  (=  AO') :  CO'- 


CB::CB:  CO', 
CB  {=  BO'). 


*  The  bieector  of  the  exterior  nn«,'le  nuetf  the  side  prodMed ;  but  in  higher  geometry,  as  it  i^i 
always  understood  that  lines  are  iudefiuite  unless  limited  by  hypothesis,  such  specifications  are 
deemed  unnecessary. 


.:^-;  /^' 


312  DsTEODUCTION  TO   MODERN   GEOMETRY. 

i'jy'Kl  OGo.  Con,  1. — Conversely,  If  a  line  AO'  be  cut  liarmonically  at 
V-       b  and  B,  and  either  of  the  harmonic  means  be  bisected^  as  AB  at  c.  the 
three  segments  CO,  CB,  CO'  icill  be  in  geometric  proportion. 

For, since  AO' :  BO' : :  AO  :  BO,  AO'  +  BO'  :  AO'  -  BO' :  :  AO  +  BO  :  AO  -  BO, 
or  2C0' :  2CB  :  :  2CB  :  OCO,  and  CO' :  CB  :  :  CB  :  CO. 

066.  Cor.  2. — lu  a  given  line,  as  ab»  as  o  approaches  the  centre 
C,  0'  recedes,  and  when  0  is  at  c,  0'  is  at  infinity,  since  CO'  =  ^• 


067-  TJieo. — The  geometric  mean  between  two  lines  is  also  the 

geometric  mean    between    their  arith- 
metic and  harmonic  means. 

Dem.— Let  AO'  and  BO'  be  the  two  lines. 
Ou  their  difference,  AB,  draw  a  semicircle, 
draw  the  tangent  O'T  and  let  fall  the  perpen- 
dicular TO.  Then  O  and  O'  are  harmonic  con- 
jugates, since  CO  x  CO'  =  CB'  (?),  CO'  is  the 

arithmetic  mean  (that  is,  \  the  sum)  of  AO'  and  BO'  (?)  and  TO'  is  tlie  geometric 

mean  (?).   .'.  CO' : TO' :  : TO' :  00'  (?). 

Queries. — Which  is  the  greatest,  in  general,  the  arithmetic,  geometric,  or  har- 
monic mean  between  two  quantities  ?    Are  they  ever  equal? 

968.  ScH. — This  proposition  affords  a  ready  method  of  finding  either  of  the 
harmonic  conjugates  O  or  O',  when  the  other  is  given.  The  student  will  show 
how. 

\^  060.  Cor.  1. — The  rectangle  of  the  harmonic  means  and  the  S2im 
of  the  extremes f  is  equivalent  to  twice  the  rectangle  of  the  extremes. 

For, CO'  X  00'  =  TO"'-  -  AO'  x  BO',  whence  2C0'  x  00'  =  2A0'  x  BO';  and, 
since  2C0'  =r  AO'  +  BO',  (AO'  +  BO')  x  00'  =  2A0'  x  BO'. 

"v.  070.  Cor.  2. — Tlie  rectangle  of  the  harmonic  mean  and  the  dif- 
ference of  the  differences  of  the  \d  and  2nd,  and  the  2nd  and  3rd, 
is  equivalent  to  twice  the  rectangle  of  these  differences. 

That  is,  00'  x  [(AC  -  00')  -  (00'  -  BO')]  =  2  (AO'  -  00')  (00'  -  BO'), 
or  00'  X  (AO  —  BO)  =  2A0  x  BO.     Let  the  student  give  the  proof. 

^  071.  Cor.  3. — If  three  quantities  are  in  harmonic  proportion 
their  reciprocals  are  in  arithmetic  proportion  {i.e.,  the  difference  be- 
tween the  1st  and  2nd  equals  the  difference  between  the  2nd  and  3rd). 

For,  from  AO',  00',  BO, we  have  the  reciprocals  ^^»  =r^/  ^^,.    Now 

J 1_   _   AO'  -   00'  _         AQ  J !_  _  00'  -  BO' 

00'         AO'  ~  00'    X    AO^  ~  00'  X  AO'  '  BO'         00' '~  00'    x    BO' 


HARMONIC    PENCILS. 


313 


BO 


But 

1 
"  00' 


AO 


BO 


OO'xAO' 


OO'xBO' 


AO 
AO' 


BO 
BO' 


(?). 


1 
00' 


-    OO'xBO" 

1 1_ 

AO'  ~  BO' 

072,  JProb, — Give7i  the  harmonic  mean  and  the  difference  be- 
tween the  extremes,  to  find  the  extremes. 

SuG's.— We  have  00'  and  fiiB,{Fig.  493,  Art.  967)  given.  Then  CO  x  CO' 
=  CT'  =r  iAB",  and  CO'  -  CO  =  00',  whence  CO'"  -  00'  x  CO'  =  iAB". 
From  this  equation  CO'  can  be  constructed  {832)y  and  the  problem  solved. 


073*  Theo, —  When  two  circles  cut  each  other  orthogonally  (i.  e., 
so  thiit  the  tangents  at  the  common  point  are  at  right  angles),  any 
line  passijig  through  the  centre  of  one, 
and  cutting  the  other,  is  divided  har- 
moniccdly  ly  the  circumferences. 

Dem.— The  tangents  being  perpendicular 
to  each  other  pass  through  the  centres, 
hence  CO  x  CO'  =  CT'.  But  CB  -  CT. 
Therefore  AO'  is  cut  hai-monically. 

Fig.  494. 


974.  Proh, — To  find  the  altitude  of  a  triangle  in  terms  of  th 
radii  of  the  escribed  circles  touching  the  adjacent  sides. 

Solution. — Let  r  and  r'  be  the  radii  of  the 
escribed  circles,  and  p  the  altitude.  Now  RT, 
AT,  and  QT  are  in  harmonic  proportion  ;  since, 
considering  the  triangle  ACT,  CQ  bisects  its  in- 
terior and  RC  its  exterior  angle  (?),  we  have 
QT  :  QA  : :,  RT  :  RA.  But  r,  p,  r',  sustain  the 
same  relation  to  each  other  as  RT,  AT,  QT; 
hence  ?•,  p,  r'  are  in  harmonical  proportion. 

Therefore,  by  ^G9)  p  (r+r')  =2rr';  or  ;?=  -?!ll. 


m 


Fig.  495. 


e.6 


>-  HARMONIC  PENCILS. 

£>7tT.  Def. — A  JPencil  of  lines  is  a  series  of  lines  diverging 
from  a  common  point. 

Def. — A  Harmonic  JPeiicil  is  a  pencil  of  four  lines  cutting 
another  line  harmonically. 

III.— In  the  following  figure  OA,  OB,  OC,  OD  constitute  a  Ediinonic  Pencil, 
if  they  divide  the  line  mn  harmonically  at  A,  B,  C,  D. 


314 


INTRODUCTION   TO   MODERN   GEOMETRY. 


976.  Tlieo. — A  harmonic  pencil  divides  harmonically  every 
line  which  cuts  it. 

Dem.— OA,  OB,  OC,  OD  being  a  harmonic  pencil,  that  is,  AD,  BD,  CD,  being 

in  harmonic  proportion,  A'D',  any  other  line 
cutting  the  pencil,  is  divided  harmonically,  so 
that  A'D',  B'D',  CD',  are  in  harmonic  propor- 
tion. Through  C  and  C  draw  parallels  to  OA, 
as  LK  and  L'K'.  Now,  from  similar  triangles, 
AB  :  BC  :  :  AO  :  CK,  and  AO  :  CL  :  :  AD  :  CD. 
But  AD  :  CD  :  :  AB  :  BC,  since  AD  is  harmoni- 
cally divided.  Hence  AO  :  CK  : :  AO  :  CL,  and 
CK  =  CL.  Hence  from  similar  triangles 
CK'  =  CL'.  Again  A'B' :  B'C  : :  A'O  :  CK'  (?), 
Fig.  49G.  and  A'D'  :  CD' : :  A'O  :  CL'  {=  CK')  (?),  whence 

A'B'  :  B'C  : :  A'D'  :  CD',  or  A'D',  B'D',  CD) 
are  in  harmonic  proportion. 

Sen. — If  the  line  through  C  cut  the  prolongation  of  AO  beyond  O,  it  is  still 
harmonically  divided ;  and,  in  fact,  it  is  scarcely  necessary  to  make  this  state- 
ment, since  in  all  general  discussions  lines  are  to  be  considered  indefinite,  un- 
less limited  by  hypothesis. 

977.  Def. — The  alternate  legs  of  a  harmonic  pencil  are  called 
conjugate,  as  OA  and  OC,  OB  and  OD. 


\d7^.  Theo. — If  tico  conjugate  legs  of  a  harmonic  imicil  he  at 
right  angles,  one  of  them  bisects  the  angle  included  by  the  other  ixiir, 
and  the  other  the  suyplement  of  this  angle, 

SxjG.— This  is  the  converse  of  {962),    remembering  that  the  bisectors  of  two 
adjacent  supplemental  angles  are  at  right  angles. 


SECTION  VI. 

ANHARMOMC    RATIO. 

979.  Di^F.—TJie  Anhavmonic  Batio  of  fonr  points  in  a 
right  line  is  the  ratio  of  the  rectangle  of  the  distance  between  the 
first  and  fourth  into  the  distance  between  the  second  and  third  to 
the  rectangle  of  the  distance  between  the  first  and  second  into  the 

distance  between  th§j^^^wi^  and  fourth. 

III. — The  anharmonic  ratio  of  the  four  points 
A,  B,  C,  D  is  AD  x  BC  :  AB  x  CD. 


B 
Fig. 


497 


AN    HARMONIC   RATIO.  315 

980.  The  relation  AD  x  bc  :  AB  x  CD  is  expressed  for  brevity 
[abcd]. 

III.— Thus  [ABCD]  means  AD  x  BC  :  AB  x  CD  ;  [ADBC]  means  AC  x  DB: 
AD  X  BC;    [BACD]   means  BD  x  AC  :  BA  x  CD;  etc.     The  ratio  [ABCD],  or 

AD  X  BC  :  AB  X  CD  is  evidently  the  same  as  ^^^   •'  ^^^r^. 

dC       \*\j 

981.  ScH. — The   appropriateness  of  the  term  anharmonic   (;^^^harmonic) 

will  be  seen  when  we  observe  that,  if  AD  is  harmonically  (\\\\(\it(\,  ^^  equals  j:—-. 

dC  CO 

If,  therefore,  g^  is  not  equal  to  ^,  which  is  the  general  case  of  division,  irre- 
spective of  the  position  of  the  points  B  and  C,  we  may  consider  the  ratio  of 
BC  ^^CD'  '^^^^  general  ratio,  or,  what  is  the  same  thing,  AD  x  BC  :  AB  x  CD, 
is  called  the  anharmonic  ratio. 


982.  Theo. — Tlie  anharmonic  ratio  of  four  x>oints  is  not  clianged 
by  interchanging  tiuo  of  the  letters,  2^^'ovided  the  other  two  be  inter- 
changed at  the  same  time. 

Dem.  [ABCD]  =  [DCBA]  =  [BADC]  =  [CDAB],  /.  e.,  AD  x  BC  :  AB  x  CD 
=  DA  X  CB  :  DC  X  BA  =  BC  x  AD  :  BA  x  DC  =  CB  x  DA  :  CD  x  AB,  which 
are  evidently  identical.  [The  student  should  notice  the  different  segments  of 
the  line  indicated  by  the  different  forms.] 

983.  ScH.— But  [ACBD]  is  a  different  anharmonic  ratio  from  [ABCD] ;  since 
AD  X  CB  :  AC  X  BD  is  not  necessarily  equal  to  AD  x  CB  :  AB  x  CD.  Now,  as 
there  can  be  twenty-four  permutations  of  four  letters,  there  may  be  formed  six 
different  anharmonic  ratios  from  four  given  points  in  a  line. 


984z.  TJieo. — If  a  ^^encil  of  four  lines  is  cut  hy  any  transversal, 
the  anharmonic  ratio  of  the  four  points  of  intersection 
is  constant. 

Dem.    SL,  SM,  SN,  SO,  or,  as  we  may  read  it,  S-L,M,N,0, 
being  such  a  pencil,  and  AD  any  transversal,  draw  through 
C  NP  parallel  to  SO.    Then, 
AD   X   BC:AB  X  CD::  jf^'^^h:  -;^N:AS)     ^^^^^p, 

<  AD  :  CD  J         ^  AS : CP ) 
But  CN  :  CP  is  constant  for  all  positions  of  C  on  SM.    There- 
fore AD  X  BC:  AB  X  CD  is  constant  for  any  transversal. 

985.  Sen. — Other  constant  ratios  may  be  written  from  the  preceding  prop- 
osition and  scholium.  Tlic  anharmonic  ratio  [ABCD]  is  called  the  anharmonic 
ratio  of  the  pencil.  The  angles  of  the  pencil  are  the  six  angles  included  by  the 
rays. 


316 


INTRODUCTION   TO   MODERN   GEOMETRY. 


086. — Cor.  1. — //'  tiuo  2)C7icils  are  mutuaUy  equiangular  their 
anharmonic  ratios  arc  equal. 

Query. — Is  the  converse  of  this  corollary  true  ?  IV.^  • 

987.  CoR.  2. — If  two  pencils  have  their  ijiter- 
sections  in  the  same  right  line,  their  anharmonic 
ratios  are  equal. 


Fig.  499. 


988.    Def. — The  anharmonic  ratio   of  four 
points  on  tlie  circumference  of  a  circle  is  the  anharmonic  ratio  of 
the  pencil  formed  by  joining  these  points  Avith  any 
point  in  the  circumference. 

III. — Thus,  the  anharmonic  ratio  of  the  points  A,  B,  C,  D  is 
the  anharmonic  ratio  of  the  pencil  0-A,B.C,D,  it  being  im- 
material where  in  the  circumfei-ence  the  point  O  is  taken, 
since  by  Cor.  1,  preceding,  the  ratio  is  the  same  for  any  posi- 
tion of  O  (?). 


BC 


Fig.  500. 


989.  TJieo. — If  four  fixed  tangents  to  a  circle  are  cut  dy  a  fifth, 

the  anharmonic  ratio  of  the  four  ])oints 
of  intersection,  called  the  anharmonic 
ratio  of  the  tangents,  is  constant. 


— V 


Fig.  501. 


Dem.  a,  B,  C,  D  being  the  fixed  points  of 
tangenc}'',  any  transversal,  as  TV,  cutting  the 
tangents,  has  the  anharmonic  ratio  [LMNP] 
constant.  For  the  pencil  0-L,IVI,N,P  lias  its 
angles  constant.  Thus  LOM  is  measured  by  \  arc  (AX — BX)  =  ^AB,  which  is  con- 
stant. And  in  like  manner  MON  is  measured  b}'-  \  arc  BC,  and  NOP  is  measured 
by  I  arc  CD.  Hence,  by  the  first  of  the  preceding  corollaries,  the  anharmonic 
ratio  [LMNP]  is  constant. 


990.  The  theory  of  anharmonic  ratio  is  applied  with  great  facility 
to  the  demonstration  of  theorems  showing  that  several  points  are 
in  a  right  line,  and  that  several  lines  intersect  in  a  common  point. 
We  give  three  specimens  of  each  class. 

991.  TJieo. — If  tico  pencils  have  the  same  anharmonic  ratio 

and  a  homologous  rag  common,  the 
intersection  of  the  other  homolo- 
gous   rays    are    in    the    same   right 

line. 

Dem.— Let  S-A,B,C,D  and    S'-A,B',C',D' 

Pjq  5Q.2.  ^^  ^^'<^  pencils  having  the  same  anharmonic 


AN   HARMONIC  RATIO. 


317 


.ratio,  and  the  rays  SA,S'A  coincident;  then  the  intersections  E,F,  H  are  in  the 
same  right  line.  Let  the  line  passing  through  E  and  F  intersect  SA  iu  K,  and  sup- 
pose it  intersect  SD  in  H',  and  S'D'  in  H".  Then,  since  the  anharmonic  ratios 
of  the  two  pencils  are  equal  [KEFH']  =  [KEFH"];  whence  H'  and  H"  are  the 
same  point,  and  must  be  the  intersection  of  the  two  lines  SD,  S'D',  that  is,  H. 

992.  Theo.—  If  in  two  right  tines  four  poi?its  in  the  one  have 
the  same  anharmonio  ratio  as  four  points  in  the  other,  and  one  ho- 
mologo2ts  2)oint  in  common^  the  three  lines X)0.8sing  through  the  other 
pairs  of  homologous  points  meet  in  a  common 
point. 

Dem.— Let  A  be  common,  and  [ABCD]  =  [A  6'C'D'J. 
Draw  SA  and  SD'.  Call  the  point  in  which  SD' 
cuts  AL  D"  (for  the  time  being).  Then  [AB'C'D'J 
=  [ABCD"].  But  by  hypothesis  [AB'C'D'J  =  [ABCD]. 
Therefore  [ABCD]  =  [ABCD"],  and  D  and  D"  are  one 
and  the  same  point.  Hence  the  three  lines  which 
pass  through  B  and  B',  C  and  C,  D  and  D'  meet  in  a 
common  point  S. 


Fig.  503. 


993.  TJieo. — If  the  lines  p>assing  through  the  corresponding  ver- 
tices oftiuo  triangles  meet  in  a  common  point,  the  intersections  of  their 
homologous  sides  lie  in  the  same  right  line. 

Dem.— Let  ABC  and  A'B'C  be 
two  triangles  so  situated  that  the 
lines  AA',  BB',  CC  meet  iu  the  com- 
mon point  S ;  then  L,  M,  N,  the  in- 
tersections of  the  homologous  sides, 
are  in  a  right  line.  For  the  pencil 
S-L,  B,  A,  C  being  cut  by  the 
two  transversals  LD,  LD',  gives 
[LBAD]  =  [LB'A'D']  {984:).  But 
C-L,B,A,D,  and  C'-UB',A',D',  have 
these  anharmonic  ratios,  hence  C-L, 
0,M,N,  and  C'-L,0,M,N,  their  equiv- 
alents, tmd  having  a  common  ray 
'CC',have  equal  anharmonic  ratios, 
and  consequently  L,  M,  N  are  in  the 
same  right  line  {991).  Fig.  m\. 


\ 


^  J  994.  TJieo. — If  the  intersection  of  the  correspo7iding  sides  of  two 
triangles  are  in  the  same  right  line,  the  lines  jjassing  through  their 
corres2)07iding  angles  meet  in  a  common  2)oint. 

Dem.— In  the  last  figure,  if  AB  and  A'B',  AC  and  A'C,  BC  and  B'C  have  their 


318 


INTRODUCTION   TO   MODERN   GEOMETRY. 


intersections  in  tlie  same  right  line,  as  LN,  the  lines  passing  through  B  and  B', 
A  and  A',  C  and  C  meet  in  a  common  point,  as  S.  By  {987)  C-L,0,M,N  has  the 
same  anharmonic  ratio  as  C'-L,0,M,N,  ^vhence  [LBAD]  =  [LB'A'D'],  and  the 
truth  of  the  theorem  follows  from  (992). 


OOo.  Theo. — The  opposUc  sides  of  an  inscriled  hexagon  have 
their  intersections  in  the  same  straight  line. 

Dem. — The  anharmonic  ratios  of  the 
pencils  B-A,E,D,C,*  and  F-A,E,D,C  be- 
ing equal  {988),  LGDE,  which  intersects 
the  first,  is  divided  in  the  same  anhar- 
monic ratio  as  NHDC,  which  cuts  the  sec- 
ond, or  [LGDE]  =  [NCDH].  But  these 
lines  have  a  common  homologous  point  D, 
hence  the  lines  joining  the  other  pairs  of 
homologous  points,  as  LN,  CC,  EH,  meet 
in  a  common  point,  as  M.  Therefore 
L,  M,  N  are  in  the  same  right  line. 

996.  Sen. — This  theorem  is  due  to' 
Pascal,  whose  wonderful  achievements 
in  his  brief  life  of  thirty-nine  years  (1623- 
1662)  have  been  the  admiration  of  all  suc- 
ceeding generations. 

Fig.  505. 


907.  Theo. 


Fig.  506. 


The  diago7ials  joining  the  opposite  vertices  of  a  cir- 
cumscribed hexagon  intersect  in  a 
common  ptoint. 

Dem.— Consider  AB.  BC,  CD,  EF  four 
fixed  tangents  cut  by  ED  and  FA.  Then 
[PNDE]  =  [AQMF]  {989).  Hence  the  an- 
harmonic ratios  of  B-P,N,D,E,*  and 
C-A,Q,M,F  are  equal  (985);  and  since 
they  have  a  common  ray  (CQ,  BN)  the  in- 
tersections A,  0,  D,  of  their  homologous 
raj's,  are  in  the  same  right  line.  Therefore 
the  diagonals  pass  through  a  common 
point. 


*  The  student  can  conceive  the  rays  BE,  BDi  elc„  as  drawn,  without  encumbering  the  figure 
with  them. 


POLE  AND   POLAR  IN  RESPECT  TO  A  CIRCLE. 


319 


SECTION  VIL 


POLE  AND  POLAR  IN  RESPECT  TO  A  CIRCLE. 

^^S.  Def. — If  a  secant  to  a  circle  be  revolved  about  a  fixed  point 
in  the  plane  of  the  circle,  the  locus 
of  the  harmonic  conjugate  of  the 
fixed  point,  in  reference  to  the  in- 
tersections, is  the  I^olav  of  the  fixed 
point.  The  fixed  point  is  the  JPole 
of  the  Polar  Line.  The  terms  pole 
and  polar  as  here  used  are  correlative 
and  neither  has  any  significance 
without  the  other. 

III. — Let  AP  be  a  secant  revolving  about 
tke  fixed  point  P,  and  let  C  be  so  taken  that 
(in  every  position)  AC  :  CB  : :  AP  •  BP,  then 

is  the  locus  of  C  the  Polar  of  P,  and  P  is  Fio.  507, 

the  Pole  of  the  locus  of  C. 


yOO'  Tlieo. — Tlie  Polar  of  a  given  point  in  respect  to  a  circle  is  a 
rigid  line. 

Dem. — Let  P  be  the  pole,  AP  any  secant 

passing  through  P,  and  a  point  in  the  polar. 

The  locus  of  C  is  required.  Draw  PL  through 

the  centre,  and  let  fall   the  perpendicular 

CC.    Draw  AL,  AH,  ACT,  and  CB.     Since 

AC  :CB::  AP:  BP,    C'P  bisects    the    angle 

BC'F,    the   exterior   angle  of   the    triangle 

^^'     AC'B  (?) ;  hence,  as  LAH  is  a  right  angle,  AL 

(^./  bisects  NAC,  the  exterior  kngle  of  the  tri- 

'  angle  CAP  (?).  Therefore,  PL  is  harmonically 

divided  at  C,  and  H  ;  and,  C  being  a  fixed 

'point,  and  C  any  point  in  the  locus,  the  locus 

is  the  perpendicular  TCCV. 


Fig.  508. 


1000.  Cor.  1. — Si7ice,  as  the  secant  revolves,  the  points  a  a?id  B 
will  vanish  in  C",  C"  is  the  point  at  which  a  tangent  from  the  pole  p 
touches  the  circle. 


1001.  CoR.  2.—Draioing  OC",    C'P,    we  see   that  OC"'  (or  oh') 
=  0C  X  OP. 


320 


INTRODUCTION   TO   MODERN   GEOMETRY. 


1002.  CoR.  3. — The  jjolar  of  a  point  hi  ilic  circumference  is  a 
tangent  at  that  point. 

For,  as  OC  x  OP  is  constant  and  equal  to  OH",  OP  diminislies  as  OC  in- 
creases, and  when  OP=OH,OC'=:OH  also. 


1003.  JProh. — To  draiv  the  ptolar  to  a  givenp)ole  in  respect  to  a 
given  circle. 

Cor.  1  effects  the  solution. 


1004.  JProb. — To  find  the  piole  of  a  polar  to  a  given  circle. 

Through  the  centre  draw  a  perpendicular  to  the  polar.     [The  student  should 
be  able  to  complete  the  solution.] 


1005.  Def. — The  point  c  wliere  the  polar  cuts  the  line  passing 
through  the  pole  and  the  centre  of  the  circle  is  called  the  Polar 
Point. 

1006.  TJieo. — The  pole  and  polar 
2mnt  are  inter  changeable. 

Dem. — TV  being  the  polar  to  P,  we  are 
to  show  that  T'V,  parallel  to  TV  and  pass- 
]15=:^>^3^  ing  through  P,  is  polar  to  C ;  i.e.^  that  any 
^""^cant,  as  AC'C",  passing  through  C,  is  di- 
vided harmonically  in  the  intersections  with 
the  circumference,  C,  and  the  intersection 
with  T'V.     Drawing   AP,  since  P  is  the 
pole  of  TV,  we  have,  as  in  the  last  demon- 
stration, angle  APB  bisected  by  PC ;  and 
consequently  RPB  bisected  by  PC".     Therefore  AC  :  CB : :  AC":  BC".  Q.  E.  D. 


Fig.  509. 


10011  *  Tlieo. — The  polars  of  all  the  points  in  a  right  line  pass 
through  the  pole  of  that  line;  and,  con- 
versely, The  poles  of  all  straight  lines  luhich 
pass  through  a  given  point  are  in  the  polar 
of  tliat  point. 

Dem. — 1st.  TV  being  a  given  line  and  P  its  pole, 
we  are  to  show  that  the  polar  to  an}'  point,  as  N , 
passes  through  P.  Draw  through  P  a  perpendicu- 
lar to  ON  ;  then  P'  is  the  polar  point  to  N. 
For,  OP  :  ON  :  :  OP' :  OB  (?) :  whence  ON  x  OP' 
=  OPxOB  =  OA^-  Therefore,  T'V  is  the  polar 
of    N  (?).      2nd.    P    being    any    point     and    TV 


Fig.  510. 


RECIPROCAL  POLARS. 


321 


its  polar,  the  pole  of  any  line,  as  T'V  passini^  through   P,  is  in  TV,  as  at  N. 
Draw  ON  perpendicular  to  T'V.    Then,  as  before, 
ON  X  OP  =  OP  X  0B=  OA^  and   N  is  the  pole  of 
TV. 

lOOS.  Cor. — The  j^ole  of  a  straight  line 
is  the  intersection  of  the  ])olars  of  any  tivo 
of  its  lioints  ;  and,  conversely,  The  i^olar  of 
any  point  is  the  straight  line  joining  the 
j)oles  of  any  tivo  straight  lines  passing 
through  that  point. 


RECIPROCAL  POLARS. 


a 


^ 


1009 .  Def. — If  two  polygons  be  constructed,  one  within,  or 
inscribed  in,  a  circle,  and  the 
other  without,  or  circumscribed 
about  the  same  circle,  such  that 
the  vertices  of  the  one  are  the 
poles  of  the  sides  of  the  other, 
the  two  polygons  are  called 
Hecii^rocal  Polars ;  and 
the  circle  is  called  the  Auxiliary 
Circle. 

The  possibility  of  constructing  such 
polygons  is  apparent  from  the  last  the- 
orem. When  the  points  P,  P',  P",  P'"  Fig,  51o 
are  in  the   circumference,  TV,  TT', 
T'V,  VV  become  tangents,  as  appears  from  (1002). 

1010.  JProb. — Raving  given  one  of  two  reciprocal polars,  to  con- 
struct the  other. 

The  student  should  be  able  to  make  the  construction. 


1011.  By  means  of  the  relation  between  reciprocal  polars  a  large 
class  of  propositions  relating  to  the  relative  positions  of  lines  and 
points,  become,  as  it  were,  double  ;  i.e.,  one  proposition  being  proved, 
another  can  be  inferred.  The  process  by  which  the  inference  is  made 
is  called  reciprocation.    We  will  give  an  example. 


322 


INTRODUCTION   TO  MOLELN   GEOMETE: 


1012.  JProb. — To  deduce  the   reciprocal    of  PascaVs   theorem 
(995). 


Solution. — Draw  tangents  at  the  six  ver. 
tices  of  the  inscribed  hexagon.  Thus,  a  cir- 
cumscribed hexagon  is  formed  whose  sides 
are  the  polars  of  the  vertices  of  the  inscribed 
hexagon,  through  the  vertices  of  which  they 
respective]}'  pass  (1002).  Now,  drawing  the 
diagonals  PM,  NQ,  OL,  they  are  the  polars 
of  the  intersections  of  the  opposite  sides  of 
the  inscribed  hexagon,  as  PM,  polar  to  the 
intersection  of  DE  and  CB  (?);  and  hence 
they  pass  through  a  common  point,  as  V. 
Thus  we  have  Brianchon's  theorem,  viz.  ; 
The  lines  joining  the  opposiie  angles  of  a  cir- 
cu7nscn'bed  hexagon  pass  through  a  common 
point. 


Fig.  513. 


1013. — The  three  following  theorems  are  of  frequent  use  in  ap- 
plying the  theory  of  reciprocation. 


1014.  TJieo.  —  The  angle  included  by  tiuo 
straight  lines  is  equal  to  the  angle  included  hy  the 
lines  joining  their  2Joles  to  the  centre  of  the  auxiliary 
circle. 

Dem. — The  pole  of  a  line  being  in  the  perpendicular  from 
the  centre  of  the  auxiliaiy  circle  upon  the  line  (1000),  O'  is 
the  supplement  of  O ;  hence  o  =  0. 


Fig.  514. 


WIS.  T/ieo. 


Tlie  distances  of  any  tiuo  points  from  the  centre 
of  the  auxiliary  circle  are  to  each  other  as  the 
distances  of  each  point  from  the  polar  of  the 
other. 

Dem.     P,  P'  being  the  points,  and  TV,  T'V  their 
polars  respectively,  we  are  to  show  that 
CP  :  CP' : :  PD"  :  P'D". 

By  UOOl)  R'  =  CP  X  CD  =  CP'  x  CD',  R  bemg 
the  radius  of  the  auxiliary  circle.    Whence 
CP  :  CP' : :  CD'  :  CD.     But  CP  :  CP' : :  CF  :  CE  (?),  and 
there  follows  CF  :  CE  :  :  CD' :  CD,  CD'  -  CF  (=  PD") 
:  CD  -  CE  (=  P'D'") : :  CF  :  CE  : :  CP  :  CP. 


ScH.— -This  is  known  as  Salmon's  Theorem. 


RADICAL  AXIS  AND   CENTRES   OF  SIMILITUDE   OF  CIRCLES.      323 

1016.  Theo. — The  anharmonic  ratio  of  four  points  in  a  straight 

line  is  equal  to  that  of  the  pencil  formed  by  the 
four  polars  of  these  points. 


Dem. — 1st.  The  polars  pass  through  a  common  point 
and  thus  form  a  pencil  (?).  2cl.  The  angles  included 
by  the  lines  joining  the  four  points  with  the  centre,  and 
those  included  by  the  polars  are  equal  (?),  hence  the 
two  pencils  have  the  same  anharmonic  ratio. 


Fig.  516. 


SECTION  VIII. 

RADICAL  AXES  AND  CENTRES  OF  SBIILITUDE  OF  CIRCLES. 

1017.  Def. — The  JPower  of  a  I^oint  in  the  plane  of  a  circle 
is  the  rectangle  of  the  distances  from  the  point  to  the  intersections  of 
the  circumference  by  a  line  passing  through  the 

point.  ^ — .A . 

III. — Thus,  the  power  of  a  point  P,  without  the  circle, 
is  PA  X  PB  ;— the  power  of  a  point  within,  as  P',  is 
P'A'  X  P'B' ;  the  power  of  a  point  in  the  circumference 
is  zerOy  since  one  of  the  distances  is  then  0 ; — the  power 
of  the  centre  is  the  square  of  the  radius.  yiq.  517. 

1018.  Cor. — Tlie  2)0iver  of  a  given  p>oint  zuith  respect  to  a  given 
circle  is  a  constant  quantity. 

Thus  PA  X  PB  =  the  square  of  the  tangent  from  P  to  the  circle,  in  whatever 
position  PB  lies,  so  long  as  it  passes  through  P.  So  also  P'A'  x  P'B'  =  the 
rectangle  of  the  segments  of  any  other  chord  passing  through  P'. 


1019.  T)E¥.—The  Hadical  Axis  of  Two  Circles  is  the  locus 
of  the  point  whose  powers  ^vitb  respect  to  tlie  two  circles  are  equal. 


324  INTRODUCTION  TO    MODERN   GEOMETRY. 

1020.  I*rojh — Tlte  Jiadical  Axis  of  two  circles  is  aright  line. 


Fig   518. 


Dem. — Join  the  centres  of  the  two  circles  O,  0',  and  take  a  point  R  on  this 
line  such  that  OR^  -  OR'  =  OT"  -  OT',  or  OR*'  -  OT'  =  OR^  -  OT'',  and 
erect  PR  perpendicular  to  00'.  Then  P  being  ariy  point  in  this  perpendicular, 
OF*  —  or'  =  OP'  —  or?'.  Adding  this  to  the  preceding  equation,  we  have 
OP-  -  or"  =  O^P'  -  0T'^  or  PT^  =  PT'".  .-.  PT  =  P T'.  PT  and  PT'  being 
tangents  to  the  circles  from  any  point  in  PR.  Hence  PV  is  the  radical  axis  of 
the  two  circles. 

1021.  Cor. —  WIie7i  the  circles  are  exterior  to  each  other,  the  Radi- 
cal Axis  lies  between  them,  touching  neither  ;  tuhen  they  are  tangeni, 
either  externally  or  inter7ially,  the  radical  axis  is  the  common  tan- 
gent J  when  they  cut  each  other,  the  axis  ^  the  common  chord  produced, 

1022»  ScH. — When  the  circles  intersect  it  might  seem  that  the  above  de- 
monstration fails  for  points  within,  as  in  the  common  chord.  But,  the  powers 
of  any  point  in  this  chord  are  still  equal.  Thus,  at  the  intersections  the  powers 
are  zero ;  and  at  any  other  point  in  the  chord,  as  a,  ab  x  ae  =  ad  x  ac,  since 
each  is  equal  to  ao  x  as. 

1023.  Cor.  There  ^  an  infinite  number  of  circles  having  their 
cejitres  in  the  same  right  line,  luhich  have  the  same  radical  axis  as 
any  two  given  circles. 

Thus,  in  the  first  figure,  PV  being  the  radical  axis  of  the  circles  O,  0',  letting 
circle  0  remain  fixed,  0'  may  vary  indefinitely  so  that  O'R^  —  O'T''  remains 
constant,  and  equal  to  OR    —  OT  . 


1024.  I*roh. — Given  two  circles,  to  draw  their  radical  axis. 

Solution. — Draw  a  common  tangent,  bisect  it,  and  through  the  point  of 
bisection  draw  a  line  perpendicular  to  the  line  joining  the  centres.  "When  the 
circles  are  tangent  to  each  other,  the  distance  between  the  points  of  tangency 
is  0 ;  hence  the  perpendicular  is  erected  at  this  point.  When  they  intersect, 
produce  the  common  chord,  or  use  the  first  method. 


CENTRES   OF   SIMILITUDE. 


325 


1025.  JProj). —  Wheji  two  circles  cut  each  other  orthogonally,  that 
is,  at  right  angles,  the  square  of  the  radius  of  either  is  equal  to  the 
poiuer  of  its  centre  luith  resj^ect  to  the  other. 


Dem.— The  power  of  0  with  respect  to  circle  O'  is 
Qa  X  On  —  OP^,  and  of  0'  with  respect  to  circle  O, 
O'b  X  O'm  =  O'P^ ;  since,  as  the  circles  cut  each  other 
orthogonally,  their  tangents  are  at  right  angles,  and  the 
tangent  to  either  passes  through  the  centre  of  the 
other. 


Fig.  519. 


1026.  IProp.—  Tlie  radical  axes  of  a  system  of  three  circles 
ivhose  centres  are  not  in  the  same  straight  line,  intersect  at  a  common 
2)oint 

v 
Dem. — Since  0,  O',  0"  are  not 
in  a  straight  line,  the  radical  axes 
of  O,  O',  and  0,  0",  as  PV"  and 
PV  intersect.  Let  P  be  their 
common  point.  Now  the  power 
of  P  with  respect  to  0'  is  equal 
to  its  power  with  respect  to  O", 
since  each  is  equal  to  its  power 
with  respect  to  O.  Hence  P  is  a 
point  in  the  radical  axis  of  O',  0". 

1027.  Cor. — If  the  centres  are  in  the  same  straight  line,  their 
radical  axes  are  parallel,  and  the  common  point  is  at  infinity. 

1028.  Per — The  intersection  of  the  radical  axes  of  three  circles 
is  called  their  Hadical  Centre, 


Fig.   5-20. 


CENTRES  OF  SIMILITUDE. 

1029.  Def. — If  the  line  joining  the  centres  of  two  circles  be 
divided  externally,  as  at  c, 
ai^d  internally,  as  at  c, 
in  the  ratio  of  the  radii, 
these  points  are  respectively 
the  External  and  the  In- 
ternal Centres  of  Sim- 
ilitude of  the  two  cir- 
cles. 

III.— If  CO  :  CO' : :  EO  :  E'O',  C  is  the  external  centre  of  similitude ;  and,  if 
CO  :  CO' : :  E^b  :  E'O',  C  is  the  internal  centre  of  similitude. 


Fig.  521. 


326  INTRODUCTION  TO  MODERN   GEOMETRY. 

The  student  should  construct  the  figure  when  the  circles  are  tangent  exter- 
nally,— when  they  are  tangent  internally, — and  when  one  is  wholly  within  the 
other. 

Query. — How  are  the  centres  of  similitude  situated  in  the  three  different 
relative  positions  of  the  circles  ? 


1030.  I^rop. — In  Uvo  circles  the  line  passing  through  the  ex- 
tremities of  two  liarallel  radii  on  the  same  side  of  the  line  passing 
through  the  centres,  intersects  this  line  in  the  external  centre  of  sim- 
ilittide,  and  if  the  radii  are  on  ojyposite  sides  of  this  line  the  inter- 
section is  the  internal  centre  of  similitude. 

The  proof  consists  in  showing  that  the  line  passing  through  the  centres  is 
divided  as  above.  Let  the  student  show  it  for  the  three  diflFerent  positions  of 
the  circles. 

1031.  Cor.  1. — Conversely,  If  any  transversal  he  draion  from 
either  centre  of  similitudey  the  radii  drawn  to  the  inter  sect  io7is  are 
parallel. 

Thus  in  the  last  figure,  since  CO  :  CO'  '•  EO  :  E'O',  and  the  triangles  have  the 
angle  C  common,  EO  and  E'O'  are  parallel. 

1032.  Cor.  2. — Tangents  draion  at  the  alternate  intersections  of 
a  transversal  through  the  external  centre  of  similitude  are  parallel ; 
also,  those  at  the  mean  intersections,  and  those  at  the  extreme  inter- 
sections, if  the  transversal  he  draion  through  the  internal  centre  of 
similitude. 

This  follows  as  a  consequence  of  the  parallelism  of  the  corresponding  radii, 
to  which  the  tangents  are  perpendicular.  Thus,  tangents  at  E  and  E'  are  par- 
allel, as  are  those  at  F  and  F'.  So,  also,  tangents  D  and  D',  and  at  E'  and  E"  are 
parallel. 

1033.  Def. — The  extremities  of  two  parallel  radii  on  the  same 
side  of  the  line  joining  the  centres  are  called  Homologous  Points, 
and  those  of  non-parallel  radii  where  the  transversal  cuts  the 
circumferences,  as  E,  F',  are  called  Anti-Homologous  Points. 

1034.  CoR.  2. — The  distances  of  a  centre  of  similitude  from  two 
homologous  points  are  to  each  other  as  the  radii. 

1035.  CoR.  3. — The  cejitres  of  similitude  and  the  centres  of  the 
circles  are  four  harmonic  points. 


1036.  JProp. — If  a  circle  touch  two  others,  the  line  joining  their 
points  of  contact  imsses  through  the  external  centre  of  similitude  of 


CENTRES    OF    SIMILITUDE.  327 

the  latter  if  the  contacts  ttre  both  external  or  both  internal ;  and 
through  the  internal  centre  of  similitude  if  the  contacts  are  the  one 
external  and  the  other  internal. 


Fig.  522. 

Dkm.— In  either  case  let  0"  be  the  circle  tangent  to  0,  and  O' ;  and  tlirouffh 
the  points  of  tangency  draw  E'C.  The  angle  0"ET  =  0"FE'  =  EFO  =  FEO ; 
whence  OE  and  O'E'  are  parallel,  and  the  similar  triangles  CEO,  CE'O'  give 
CO  :  CO'  :  :  OE  :  O'E'.    [The  student  should  make  the  otlier  constructions.] 


1037.  Froh. — To  draw  a  li7ie  2-)arallel  to  a  given  line  so  that  the 
distance  betiveen  the  extreme  intersections  with  two  ^iven  circles  shall 
be  a  maximum. 

SoLUTiON.-^Draw  a  line  through  the  internal  centre  of  similitude  and  par- 
allel to  the  given  line.  Now,  at  the  extreme  intersections  draw  tangents,  and 
it  will  become  evident  that  the  line  first  drawn  is  a  maximum.  [The  student 
should  make  the  figure  and  fill  out  the  proof] 

If  the  circles  are  wholly  exterior  to  each  other,  the  distance  between  the 
mean  intersections  is  a  minimum. 


1038.  Concluding  Note. — Our  limits  preclude  our  pursuing  these  topics 
farther.  We  have  given  enough  to  make  the  language  of  the  Modern  Geome- 
try intelligible,  and  to  afford  some  insight  into  its  character.  One  of  the  best 
elementary  resources  for  the  Euglish  student  who  wishes  to  pursue  the  subject 
at  greater  length,  is  Mulcahy's  PHnciples  of  Modern  Oeometry,  Dublin,  1862. 
It  is,  however,  much  to  be  regretted,  that  there  is  no  English  treatise  which 
presents  the  elements  of  this  subject  with  the  philosophic  elegance  of  the 
French.  The  best  of  the  latter  is  Rouche  and  Comberousse's  Treatise  on 
Ei^ementary  Geometry.  For  a  more  extended  view-  of  the  subject,  Salmon  or 
Whitworth  will  furnish  the  Euglish  student ;  but  he  who  would  be  proficient 
must  read  the  works  of  Cuasles  and  Poncelet,  who  are  the  irreat  authorities. 


3"^ 


OLNEY'S   MATHEMATICAL    SERIES. 


ELEMENTS 


OP 


TEIGON-QMETEY, 


PLANE    AND    SPHERICAL. 


Bt    EDWARD    OLNEY, 

PROFESSOR  OP   MATHEMATICS  IN   THE  rNIVIIRSlTT  Of  MICHIOAX. 


NEW    YORK: 

SHELDON    &    COMPANY, 

No.   8    MURRAY    STREET. 
1877. 


Stoddard's  lathematical  Series. 


-MM- 


STODDARD'S  JUVENILE  MENTAL  ARITHMETIC 

STODDARD'S  INTELLECTUAL  ARITHMETIC  .... 

STODDARD'S  RUDIMENTS  OF  ARITHMETIC 

STODDARD'S  NEW  PRACTICAL  ARITHMETIC        .... 

SKORT  AND   FULL  COURSE  FOR  GRADED  SCHOOLS. 

STODDARD'S  PICTORIAL  PRIMARY  ARITHMETIC 

STODDARD'S  COMBINATION  ARITHMETIC 

STODDARD'S  COMPLETE  ARITHMETIC  .... 

The  Combination  Scliool  Aritlimetic  being  Mental  and  Written  Arithmetic 
in  one  book,  will  alone  serve  for  District  Schools.  For  Academies  a  full  high 
course  is  obtained  by  the  Complete  Arithmetic  and  Intellectual  Arithmetic. 

OLNEY'S  HIGHER  MATHEMATICS. 

A  COMPLETE  SCHOOL  ALGEBRA  in  one  volume,  390  pages.  Designed 
for  Elementary  and  higher  classes  in  Schools  and  Academies.  By  Prof. 
Edward  Olney,  University  of  Michigan. 

A  GEOMETRY  AND  TRIGONOMETRY  in  one  vol.  By  Prof.  Edward 
Olney.    One  vol.  8vo,  

A  GENERAL  GEOMETRY  AND  CALCULUS  in  one  vol. 


Entered  according  to  Act  of  Congress,  in  the  year  1870, 
Br  SHELDON  A  COMPANY 
la  the  Office  of  the  Librarian  of  Congress  at  Washiu^tco. 


CONTENTS 


PART  IV.— TRIGONOMETRY. 

CHAPTER  I. 

PLANE  TRIGONOMETRY. 

SECTION  I. 
Definitions  and  Fundamental  Relations  between  the  Trigonometri- 
cal Functions  op  an  Angle  (or  Arc).  page. 

Definitions 1-5 

Fundamental  Relations  of  tlie  Trigonometrical  Functions  of  an 

Angle 5-7 

Signs  of  the  Functions 7-10 

Limiting  Values  of  the  Functions 10-1-4 

Functions  of  Negative  Arcs 15-16 

Circular  Functions 1 G 

Exercises 17-30 

SECTION  II. 
Relations  between  the  Trigonometrical  Functions  of  different  An- 
gles (or  Arcs). 

Functions  of  the  Sum  or  Difference 20-25 

Functions  of  Double  and  Half  Angles 26 

Exercises 26-29 

SECTION  III. 

Formula  for  rendering  Calculable  by  Logarithms  the  Algebraic  Sum 

of  Functions 29-30 

Exercises 30-31 

SECTION  IV. 
Construction  and  Use  of  Tables. 

Definitions 31-32 

To  Compute  a  Table  of  Natural  Functions 32-33 

To  Compute  a  Table  of  Logarithmic  Functions 33-34 

Exercises  in  the  Use  of  the  Tables 34-39 

Functions  of  Angles  near  the  Limits  of  the  Quadrant 39-42 

Exercises 43 


IV  CONTEXTS. 

SECTION  V. 

Solution  op  Plane  Triangles.  page. 

Of  Right  Angled  Triangles 44-45 

Exercises  and  Examples 45-48 

Practical  Applications 48-49 

Of  Oblique  Angled  Plane  Triangles 49-51 

Exercises 51-55 

Oblique  Triangles  Solved  by  means  of  Right  Angled  Triangles . . .  55-56 

Exercises 56-57 

Functions  of  the  Angles  in  Terms  of  the  Sides 58-59 

Exercises 59-60 

Area  of  Plane  Triangles 60-61 

Practical  Applications 61-64 


CHAPTER  11. 

SPHERICAL  TRIGONOMETRY. 

IXTRODUCTIOX. 

Projection  of  Spherical  Triangles. 

Definitions  and  Fundamental  Propositions 65-66 

Projection  of  Right  Angled  Spherical  Triangles 66-71 

Projection  of  Oblique  Angled  Spherical  Triangles 71-73 

SECTION  I. 

Solution  op  Right  Angled  SruERiCAL  Triangles. 

Definitions 73-75 

Exercises 75-76 

Napier's  Rules 76-78 

Determination  of  Species 79-81 

Exercises  in  Solution  of  Right  Angled  Spherical  Triangles 81-84 

Quadrantal  Triangles 84-85 

SECTION  XL 
Op  Oblique  Angled  Spherical  Triangles. 

To  find  the  Segments  of  a  Side  made  by  a  Perpendicular  let  fall 

from  the  opposite  angle 85-86 

The  Relation  of  the  Sides  and  Opposite  Angles 87 

Solution  of  Oblique  Spherical  Triangles  by  Napier's  Rules,  in 

Three  Problems 87-90 

Exercises •  -  90-95 


CONTENTS.  V 

SECTION  III. 

General  Formula.  page. 

Angles  as  Functions  of  Sides,  and  Sides  as  Functions  of  Angles.     96-100 

Gauss's  Equations 100-101 

Napier's  Analogies 103 

Exercises  in  the  Use  of  these  Formulae 102-107 

SECTION  IV. 

Area  of  Spherical  Triangles 108-110 

Practical  Applications  of  Spherical  Trigonometry 110-110 

TABLES. 

Introduction  to  the  Table  of  Logarithms 1-10 

Table  of  Logarithms  of  Numbers 11-28 

Table  of  Natural  Sines  and  Cosines,  and  of  Logarithmic  Sines,  Cosines. 

Tangents,  and  Cotangents 29-74 

Table  for  Precise  Calculation  of  Functions  near  their  Limits 75-78 

Table  of  Tangents  and  Cotangents 79-88 


PART    IV. 

TRIGONOMETRY, 


CHAPTER  L 

PLANE   TBIGONOMETRY. 


SECTION  L 

DEFINITIONS  AINT)  FUNDAMENTAL   RELATIONS  BET^ITIEN  THE 
TRIGONOMETRICAL  FUNCTIONS  OF  AN  ANGLE  (OR  ARC). 

1*  Trigonometry  is  a  part  of  Geometry  which  has  for  its  sub- 
ject-matter, Angles.  It  is  chiefly  occupied  in  presenting  a  scheme 
for  measuring  and  comparing  angles,  by  means  of  certain  auxiliary 
lines  called  Trigonometrical  Functions,  in  investigating  the  relations 
between  these  functions,  and  in  the  solution  of  triangles  by  means 
of  th3  relations  between  their  sides  and  the  trigonometrical  functions 
of  their  angles. 

2.  I^lane  Trif/ononietri/  treats  of  plane  angles  and  triangles^ 
in  distinction  from  Spherical  Trigonometry,  which  treats  of  spherical 
angles  and  triangles. 

3.  A  FuiictiOil  is  a  quantity,  or  a  mathematical  expression, 
conceived  as  depending  upon  some  other  quantity  or  quantities  for 
its  value. 

Ill's. — A  man's  wages /(>r  a  given  time  is  a  function  of  tlie  amount  receivetl 
per  day ;  or,  in  general,  his  wages  is  a  function  of  boili  the  time  of  service  and 
the  amount  received  per  day.  Again,  in  the  expressions  y  =  ^ax^^y  ='x*  — 
2i».T  +  5,y  =  21ogaa;,  y  =  a^,2/  is  a  function  of  a; ;  since,  the  numbers  2,  5,  a  andd 
being  considered  fixed  or  constant,  the  value  of  y  depends  upon  the  value  we 
assign  to  x.  For  a  like  reason  such  expressions  as  V^a"  —  x*,  and  Zax^  —  2>/l^^ 
may  be  spoken  of  as  functions  of  x.  Once  more,  the  area  of  a  triangle  is  a  func- 
tion of  its  base  and  altitude. 

4.  Angles  as  Functions  of  Ares. — We  have  learned  in 
Geometry  (Part  II.,  Sec.  VI.),  that  angles  and  arcs  may  be  treated 
as  functions  of  each  other;  and  that,  if  the  angles  be  taken  at  the 

1 


2  PLANE  TRIGONOMETRY. 

centre  of  the  same  or  equal  circles,  tlie  arcs  intercepted  nave  the 
same  ratio  as  the  angles  themselves,  and  hence  may  be  taken  as  their 
measures  or  representa-tives.  For  trigonometrical  purposes,  an  angle 
is  considered  as  measured  by  an  arc  struck  with  a  radius  1,  from  the 
angular  point  as  a  centre. 

o.  A  Degree  being  the  ^^  part  of  the  circumference  of  a  circle, 
becomes  the  measure  of  -5^  of  a  right  angle;  and,  for  convenience,  it 
is  customary  to  speak  of  such  an  angle  as  an  angle  of  one  degree,  of 
four  times  as  large  an  angle  as  an  angle  of  four  degrees,  etc.,  apply- 
ing the  term  directly  to  the  angle.  A  small  circle  written  at  the 
right  and  a  little  above  a  number  indicates  degrees  (°). 

6.  A  Minute  is  -^  part  of  a  degree.  Minutes  are  designated  by 
an  accent  (').  A  Second  is  -^  part  of  a  minute.  Seconds  are 
indicated  by  a  double  accent  (").  Smaller  divisions  of  angles  (or 
arcs)  are  most  conveniently  represented  as  decimals  of  a  second, 
though  the  designations  thirds,  fourths,  etc.,  are  sometimes  met  with, 
and  signify  further  subdivisions  into  60ths.  5°  12'  16"  13'"  is  read, 
"  5  degrees,  12  minutes,  16  seconds,  and  13  thirds." 

Ill's. — In  Fig.  1  AOB  is  an  angle  of  35°,  because  the  measuring  arc  db 
contains  35  of  the  360  equal  parts  mto  which 
the  circumference  whose  radius  is  Oa,  could  be 
divided.  In  like  manner  BOC  is  an  angle  of  7°. 
BOC  =  iAOB  =  iAOC-  Hence,  it  becomes  evi- 
dent that  we  may  use  the  numbers  35,  7,  and  42, 
to  represent  the  respective  angles  AOB,  BOC. 
and  AOC,  or  the  corresponding  arcs  o^,  6c,  and 
ac. 

7.  A  Ouadt^ant  is  an  arc  of  90°, 

and  is   the   measure   of  a   right  angle ; 

hence,  a  right  angle  is  called  an  angle  of 

^- 1-  90°.    Thus  arc  ad,  Fig.  1,  =  90°;  or  angle 

AOD  =  90°. 

S,  TJie  Complement  of  an  angle  or  arc  is  what  remains  after 
subtracting  the  angle  or  arc  from  90°.  The  Supplement  of  an  angle 
or  arc  is  what  remains  after  subtracting  the  angle  or  arc  from  180°. 

Ill's. — In  Fig.  1,  the  angle  BOD  is  the  complement  of  AOB,  and  the  arc  bd 
18  the  complement  of  arc  ab.  The  complement  of  35°  is  90°  —  35°=-  55°.  The 
supplement  of  35°  is  180°-  35°=  145°. 


DEFINITIONS  AND   FUNDAMENTAL  RELATIONS.  3 

,9.  A  Quadrant  is  often  represented  by  I'T,  since  -?.  is  the  semi- 
circumference  when  the  radius  is  unity.   When  this  notation  is  used, 

180° 
the  Unit  ^rc  becomes =  57°.29578  nearly,  or  57°  IT'  44".8  +, 

'71' 

which  is  an  arc  equal  in  length  to  the  radius. 

10,  For  trigonometrical  pui-poses,  an  angle  is  conceived  as  g^T. 
erated  by  the  revolution  of  a  line  about  the  angular  point,  ana 
hence  may  have  any  value  luhatever,  not  only  from  0°  to  180°,  but 
from  0°  to  360°,  and  even  to  any  number  of  degrees  greater  than 
300°,  as  1280°,  etc.  An  angle  of  45°  is  generated  by  \  of  a  revolu- 
tion, 90°  by  }  of  a  revolution,  180°  by  4  a  revolution,  270°  by  }  of  b 
revolution,  360°  by  one  revolution,  450°  by  1}  revolutions,  1280°  by 
3|  revolutions,  etc.,  etc. 

11,  In  accordance  with  the  conception  of  an  angle  as  generated 
by  a  revolving  line,  the  measuring  arc  is  considered  as  orifjinating 
at  the  first  position  of  the  revolving  line-  {i,  e.,  with  one  side  of  the 
angle),  and  terminating  in  the  line  after  it  has  generated  tlie  angle 
under  consideration  (*.  e.,  with  the  other  side  of  the  angle).  The 
first  extremity  is  called  the  Origin  of  the  arc,  and  the  other  the 
Termination. 

Ill's. — In  Fig.  1,  let  the  angle  AOB  be  considered  as  generated  by  a  line 
starting  from  the  position  OA,  and  revolving  around  the  point  0,  from  right  to 
left,*  till  it  reaches  the  position  OB.  Oa  being  taken  as  unity,  the  arc  ab  is  the 
measuring  arc  of  the  angle  AOB  ;  a  is  its  origin,  and  b  its  ierminaiion. 

12.  In  the  generation  of  angles  by  means  of  a  revolving  line,  the 
normal  motion  is  considered  to  be  from  right  to  left,  and  the  quad- 
rants are  numbered  1st,  2d,  3d,  and  4th,  in  the  order  in  which  they 
are  generated. 

13.  The  Trig onometr leal  Functions  are  eight  in  num- 
ber; viz.,  sine,  cosine,  tangent,  cotangent,  secant,  cosecant,  versed- 
sine,  and  cover sed- sine.  These  lines  are  functions  of  angles,  or. 
what  amounts  to  the  same  thing,  of  arcs  considered  as  measures  of 
angles,  and  are  the  characteristic  quantities  of  trigonometry. 

14z.  Hie  Sine  of  an  angle  (or  arc)  is  a  perpendicular  let  fall 
from  the  termination  of  the  measuring  arc  upon  the  diameter  passing 
through  the  origin  of  the  arc.  Thus  in  Fig.  2,  M  is  in  each  case 
the  sine  of  the  angle  AOB,  or  of  the  arc  axh. 

*  The  pnpil  will  understand  that,  if  he  imagines  himeelf  ptanding  at  the  centre  of  moti -u. 
as  the  moving  body  or  point  passes  before  him,  the  distinctions  "  from  right  to  left."  and 
"  from  left  to  right,"  are  easily  made. 


PLANE  TRIGONOMETRY. 


lo.  Hie  Tri{/ono2}ietrical  Tanf/ent  of  an  angle  (or  arc) 
is  a  tangent  drawn  to  the  measuring  arc  at  its  origin,  and  limited 


Fig.  2. 

by  tlie  prodrbced  diameter  passing  throngli  the  termination  of  the 
arc.  Thus  in  Fig,  S,  ac  is  in  each  case  the  tangent  of  the  angle  AOB, 
or  of  the  arc  axb. 

16,  The  Secant  of  an  angle  (or  arc)  is  the  distance  from  the 
angular  point,  or  centre  of  the  measuring  circle,  to  the  extremity  of 
the  tangent  of  the  same  angle  (or  arc).  Thus  in  Fig.  2,  Oc  is  in 
each  case  the  secant  of  the  angle  AOB,  or  of  the  arc  axb. 

17,  The  Versed-Sine  of  an  angle  (or  arc)  is  the  distance 
from  the  foot  of  the  sine  of  the  same  angle  (or  arc)  to  the  origin  of 
the  measuring  arc.  Thus,  in  Fig.  2,  da  is  in  each  case  the  versed- 
sine  of  the  angle  AOB,  or  of  the  arc  axh. 

IS.  The  prefix  co,  in  the  names  of  the  four  trigonometrical  func- 
tions in  which  it  occurs,  is  an  abbreviation  for  the  word  comjyUment. 
Thus  cosine  means  complement-sine,  i.  e.,  the  sine  of  the  comple- 
ment; cotangent  means  tangent  of  the  complement;  etc.  The  co- 
sine of  40°  is  the  sine  of  90°  —  40°,  or  50° ;  the  cosine  of  110°  is  the 
sine  of  90°  -  110°,  or  —  20°;  the  cotangent  of  30°  is  the  tangent  of 
60°  ;  the  cosecant  of  200°  is  the  secant  of  -  110°. 

19.  Construction  of  the  Coniplefuentary  Functions. 

— Let  us  now  see  how  the  complementaiy  functions  are  constructed  with  refer- 
ence to  their  primitives,  premising  that  all  arcs  in  Fig.  3,  reckoned  from  A,  are 


DEFINITIONS   AND   FUNDAMENTAL   RELATIONS. 


to  he  reckoned  around  from  right  to  left  in  this  discussion.     1st.    Let   AP    be 
any  arc  less  than  90°  ;  then  90°  —  AP  =  aP  is  its  complement.    Now  considering 
a  as  the  origin  and  P  the  termination  of 
this  complementary  arc,  Pd  is  its  sine,  at 
its  tangent,  Ot  its  secant,  and  ad  its  versed- 
sine.  Hence,  Pd^  at,  Ot,  and  ad  are  respect- 
ively the  cosine,  cotangent,  cosecant,  and 
coversed-sine  of  the  arc  AP,  or  the  angle 
AOP.  2d.  Letting  APP' be  any  arc  between 
90°  and  180°,  its  complement  is  90°  —  APP' 
or  —  aP',  the  —  sign  signifying  that  the  arc 
is  reckoned  backward  from  P'  to  a.    But  as 
the  values  of  the  functions  will  be  the  same 
whether  the  origin  be  taken  at  P'  or  at  a, 
we  may  take  a  as  the  origin  of  this  comple- 
mentary arc,  and  P'  as  its   termination, 
whence   P'c?' becomes  its  sine,  rt^'   its  tan- 
gent, 0^'  its  secant,  and  ad'  its  versed-sine. 
Therefore  P'd',  at'.  Of,  and  ad\  are  respect- 
ively the  cosine,  cotangent,  cosecant,  and  ^^^-  3- 
coversed-sine  of  the  arc  APP',  or  the  angle  AOP'.    3d.  In  like  manner,  aP"  is 
shown  to  be  the  complement  of  arc  APP'P " ;  and  as  P"d",at,  Ot,  and  ad" 
are  respectively  the  sine,  tangent,  secant,  and  versed-sine  of  this  complement, 
they  are  the  corresponding  cofunctious  of  the  arc  APP'P",  or  the  salient  angle 
AOP"*    4th.  In  the  same  way,  it  appears  that  P"'d"',  at',  Ot\  and  ad"'  are  the 
cosine,  cotangent,  cosecant,  and  coversed-sine  of   the  arc  APP'P" P",  or  the 
salient  angle  AOP"'.     Observe  that  a.^  a  point  on  the  measuring  arc  90°  from  the 
pritmlive  origin,  is  tJie  oHgin  of  all  the  complementary  fiuictions. 

ScH. — It  will  readily  appear  from  the  figure  that  the  cosine  of  an  angle 
(or  arc)  is  always  equal  to  tJie  distance  from  the  foot  of  the  sine  to  tJie  vertex  of  the 
angle  (or  the  centre  of  the  measuring  arc).  This  is  the  more  convenient  prac- 
tical definition.  Thus  the  cosine  of  AP  is  PfZ  =  DO;  the  cosine  of  APP'  is 
P'd'  =  D'O,  etc. 

20.  Notation, — Letting  x  represent  any  angle  (or  arc),  the 
several  trigonometrical  functions  of  it  are  writteL  sin  a:,  cosx,  hxnx, 
cot  a;,  sec  a;,  cosecT,  vers.T,  and  covers  a;.  They  are  read  -^sine-r," 
"  cosine  x"  "  tangent  a;,"  "  cotangent  ic,"  etc. 


FUNDAMENTAL  RELATIONS  BET^^TEN  THE   TRIOONOMETRICAI. 
FUNCTIONS  OF  AN  ANGLE  (OR  ARC). 

[Note.— These  fundamental  relations  must  he  made  perfectly  familiar.  'XTiey  must  i>e 
meaiorized,  and  be  as  familiar  as  the  Multiplication  Table.  The  student  can  do  nothing  in 
trigonometry  without  them.] 


The  discussions  in  this  treatise  all  proceed  upon  one  general 
plan;  viz., — First  obtain  the  j^ai'ticiilar  x^^^operty  of  the 


6 


PLA>iE  TRIGOXOMETEY. 


sineund  cosine,  and  from  this  deduce  all  the  others 
according  to  the  dependencies  shown  in  tlie  foUoiv- 
ing  proposition* 

21,  I*rop. — TJie  Fundamental  Relations  which  the   Trigono- 
metrical Functions  sustain  to  each  other  are: 

1 


(1)  sin' a;  +  cos' a;  =  1 


(2) 

sm  X 

tan  X  = ; 

cosa; 

(3) 

cos  a; 

cot  X  =  -. —  ; 
sma; 

(4) 

1 
cot  X  =  . ; 

cos  a; 
1 


(5)  seca;  = 

(6)  coseca: 
^  '  sma;' 

(7)  sec'  a:  =  1  +  tan'  x ; 

(8)  cosec'a;  =  1  4-  cot' a; 

(9)  vers  a;  =  1  —  cos  a; ; 
(10)  covers  a;  =  1  —  sin  a;. 


(The  forms  sin'a",  sec'a;,  etc.,  signify  tlie  square  of  the  sine,  the 
square  of  the  secant  of  x,  etc.,  and  are  read  "  sine  square  a;,"  "  secant 
square  x"  etc.  The  student  should  distinguish  between  sin'a;,  and 
gin  a;'.) 

Dem.— In  Fig.  4,  let  x  represent  any  arc  as  AP,  less  than  90°.  Then  PD  =  sin  x, 
OD  or  Pd  =  cos  J-,  AT  =  tana;,  OT  =  secx,  at  =  coti,  Ot  =  cosec  j,  AD  =  yei-sin  x, 
and  ad  =  coversin  x. 


Fig.  4. 


(1).  In  the  right-angled  triangle  POD, 
pd'  +  OD'  =  0P'>  or  sm^'a;  +  cos' a;  =  1, 
since  OP  =  radius  =  1. 

(2).  From  the  similar  triangles  POD  and 

TOA, 

AT      PD  sin  a; 

£J-  =  ~^,  or  tan  x  = 

OA      OD'  cos  a-. 

(3).  From  the  similar  triangles  POcf  and 
iOa, 

at       Pd  cos  a; 

=  ^^..  or  cot  X  —  —. —  . 

sma; 


^    =  7^'  01"  cot  X 
Oa      Od 


(4).  Multiplj-ing    (2)  and    (3)    together, 

sinajcosar      .        ^  1 

: —  =  1,  01  tan  X  —  — — 

cosajsma;  cot* 


tan  a;  cot  2-  = 


(5).  From  the  similar  triangles  OTA  asd 
CRD, 

1 


^  =  ^ ;  but  OP  and  OA  each  =  1,  .*.  sec x  = 


DEFINITIONS  AND   FUNDAMENTAL  DELATIONS.  i 

(6).  From  the  similar  triano:les  Ota  and  OPd, 

Ot       OP  1 

"TT-  =  7^,  or  cosecic  =  -; — . 
Oa      Od  sma* 

(7).  From  the  right-angled  triangle  OAT, 

or"  =  oa'  +  AT",  or  sec'jj  =  1  +  tan'a;. 

(8).  From  the  right-angled  triangle  Oat^ 

oT  =  Oa'  +  "a^y  or  cosec^^  =  1  +  cofa;. 
(9).  AD  =  AO  -  OD,  or  vers  a;  =  1  -  cos  aj. 
(10).  ad  =  aO  —  Od,  or  covers  aj  =  1  —  sin  «. 

Thus  the  fundamental  relations  of  the  functions  are  established  for  an  arc 
less  than  90°.  But  it  will  readily  a])pear  that  the  relations  are  the  same  for  any 
other  arc.  For  example,  let  x  =  AP'  be  any  arc  between  90°  and  180^  Theu 
the  triangle  P'D'O  gives  sin^'a;  +  cos'* a;  =  1,  since  P'D'  =  sin  x,  and  OD'  =  cos  x. 

The  similar  triangles  P'D'O  and  OAT'  give  =rr-  =  frj^,  or  tan  x  =  ;  and  the 

'^  *       OA        D'O'  cos  a;' 

cos  X 
similar  triangles  P'd'O  and  t'aO  give  cot  x  =  - — .    In  like  mianner  let  the 

student  observe  the  relations  when  x  =  APP'P",  or  an  arc  between  180°  and 
270°.     So  also  when  x  =  APP'P"P"'.  or  an  arc  between  270°  and  360°. 


22,  CoK.  1. — The  tangent  and  cotangent  of  the  same  angle  are 
reciprocals  of  each  other  ;  so  also  are  the  secant  and  cosine,  and  the 
cosecant  and  the  sine.     Thus,  if  tan  x  =  Z,Qotx  =  \]  since  cot  a;  = 

, .  If  sec  2;  =  2,  cos  a;  =  :|:  sinceseca;  = ,  or  cosa;  = . 

tanrc  cos  a;  seca; 

23.  Cor.  2. — Sines  and  cosines  cannot  exceed  1.  Tangents  and 
cotangents  can  have  any  values  from  0  to  ao»  Secants  and  cosecants 
can  have  any  values  hetioeen  1  and  co .  Versed-sines  and  cover sed- 
mies  can  have  any  values  'between  0  and  2.  These  conclusions  will 
readily  appear  from  the  definitions,  and  an  inspection  of  Fig,  \, 


SltJNS  OF  THE  TRIGONOMETRICAL  FUNCTIONS. 

24:,  I*rop. — Angles  (or  arcs)  considered  as  generated  from  right 
to  left  being  called  positive  *  and  marked  +,  those  considered  as  gcH" 
eratedfrom  left  to  right  are  to  be  called  negative  and  marked  — . 

•  This  is  purely  an  arbitrary  convention.    We  might  equally  well  reTerse  ii 


PLA^'E  TRIGONOMETRY. 


Dem. — Tliis  is  a  direct  application  of  the  significance  of  the  +  and  —  signs. 
Bee  Complete  School  Algebra,  pp.  20-23.)    Thus,  in  Fig.  5,  if  the  angle  AOP, 

considered  as  generated  by  the  revolution 
of  a  line  trom  the  position  OA  in  the  direc- 
tion of  the  arrow-head  (from  right  to  left), 
is  called  positive  and  marked  + ,  an  angle 
generated  by  the  motion  of  a  line  from  the 
position  OA  in  the  opposite  direction  (from 
left  to  right),  as  the  angle  AOP"  thus  gen- 
erated, is  to  be  considered  negative  and 
marked  — .  Let  it  be  carefully  observed 
that  it  is  the  assumed  direction  of  the 
motion  of  the  generatrix  that  determines 
the  sign  of  the  angle  (or  arc).  Two  lines 
meeting  at  a  common  point  may  be  con- 
sidered as  designating  either  a  -f  or  a  — 
angle,  according  to  the  direction  of  motion 
assumed.  Thus  the  lines  OA  and  OP',  Mg. 
5,  may  form  the  positive  angle  measured 
Fig.  5.  by  the  arc  APP',  or  the  negative,  salient 

angle  measured  by  the  negative  arc  AP"'P"P'.    q.  e.  d. 


2S,  JPvoj)* — Radius  being  considered  as  alioays  extending  in  the 
same  direction,  viz.,  from  the  centre  toward  the  circumference,  is 


always  positive. 


26,  JProj), — T/te  sign  of  the  sine  of  an  angle  between  0°  and  180° 
being  +,  that  of  an  angle  between  180°  and  360°  is  — . 

Dem. — In  Fig.  5,  we  observe  that  the  sines  of  all  angles  terminating  in  the  1st 
and  2d  quadrants,  i.  e.,  between  0°  and  180°,  may  be  considered  as  measured 
from  the  primarj'  diameter  AB,  npu>ard,v;\n\Q  those  of  angles  terminating  in  the 
3d  and  4th  quadrants,  i.  e.,  between  180°  and  360°,  are  reckoned  downward 
irom  the  same  line;  hence,  the  former  being  called  +,  the  latter  shojild  be  — , 
.w  t7i€  two  species  are  estimated  in  opposite  directions,    q.  e.  d. 

A  more  elegant  conception  is  to  consider  the  sine  as  projected  upon  the  diam- 
eter vertical  to  that  passing  through  the  origin,  as  aC  ;  whence  Od  is  the  sine 
of  AOP  (or  arc  AP).  Now  this  line  evidently  is  0  when  the  angle  is  0  ;  and  as  the 
angle  increases,  the  sine  increases,  being  generated  from  O  vpicard,  and  hence 
is  called  + .  This  is  the  same  conception  as  we  use  in  the  case  of  th^  cosine. 
Adopting  it,  we  see  that  sines  reckoned  from  0  upward  are  +,  and  downward 
— .    Cosines  reckoned  from  0  to  the  right,  are  + ,  and  to  the  left,  — . 

27»  Cor. — T7ie  cosecant  of  an  arc  has  the  same  sign  as  its  sine, 

aince  coseca;  =  -. — ;  and  as  1,  being  the  radius,  is  -f ,  the  sign  of 


sma; 


%\nx 


is  the  same  as  the  sign  of  sin  x. 


DEFINITIONS  AND   FUNDAMENTAL   RELATIONS.  9 

28,  IProp. — The  sign  of  the  cosine  of  an  angle  letween  0'  and 
90°,  and  between  270°  and  360°,  is  +,  lohile  that  of  an  angle  hetiucen 


Dem. — In  Fig.  5,  we  observe  that  the  cosines  of  all  angles  terminating  in  the 
1st  and  4th  quadrants,  may  be  considered  as  estimated  from  the  centre  towarc 
the  right,  as  OD,  OD'" ;  while  correspondingly,  the  cosines  of  angles  terminating 
in  the  2d  and  3d  quadrants  will  be  estimated  from  the  centre  toward  the  left,  as 
OD',  OD".  Hence,  by  reason  of  this  opposition  of  direction,  the  former  are 
called  + ,  and  the  latter  — .    q.  e.  d. 

29,  Cor. — TJie  secant  of  an  angle  has  the  same  sign  as  its  cosine, 
Biuce  these  functions  are  reciprocals  of  each  other,     (See  2*^*) 


30.  JProp. — The  sign  of  the  tangent  of  an  angle  hettveen  0°  and 
90°,  and  also  letioeen  180°  and  270°,  is  +  ;  while  that  of  an  angle 
hetiveen  90°  and  180°,  and  betiueen  270°  and  360°,  is  — . 

Dem. — Since  tan  x  = ,  when  sin  a;  and  cos  x  have  like  signs,  tan  a;  is  + ,  by 

cos  X 

the  rules  of  division ;  and  when  sin  x  and  cos  x  have  different  signs,  tan  a;  is  — . 
Now,  in  the  1st  and  3d  quadrants*  the  signs  of  sin  a;  and  cos  x  are  alike,  hence 
in  these  quadrants  tana;  is  plus;  but  in  2d  and  4th  quadrants  sin  a;  and  cos.e 
have  unlike  signs,  and  consequently  in  these  tan  x  is  — .    q.  e.  d. 

31,  Cor. — The  sign  of  the  cotangent  is  the  same  as  the  sign  of  the 

tangent  of  the  same  angle j  since  cot  x  = . 

tan  3/ 


32,  JProp, —  Versed-sine  and  coversed-sine  are  always  +. 

Dem. — Vers  a;  =  1  -  cos  a; ;  and  as  cos  a;  cannot  exceed  1,  1  — cosa;  is  al 
ways  +.  In  like  manner,  covers  a:  =  1  —  sin  a;;  and  as  sin  a;  cannot  exceed  1, 
1  —  sin  a;  is  always  + .    q.  e.  d. 

ScH.  1. — It  is  essential  that  the  law  of  the  signs,  as  explained  above,  be  well 
understood,  and  the  facts  fixed  in  memory.  Fig.  6  will  aid  the  student  in  fixing 
the  law  in  the  memoiy.  Having  this  constantly  be- 
fore the  mind,  and  remembering  that  tan  and  cot 
are  +  when  sin  and  cos  have  like  signs,  and  —  when 
they  have  unlike,  and  that  cos  and  sec  have  like 
signs,  as  also  sin  and  cosec,  or,  more  simply,  that 

1 


Cin  =  — ,  cot  =  : — ,  sec  =  — ,  and  cosec  = 

cos  tan  cos  sm 

the  student  cannot  fail  to  know  the  sign  of  a  func- 
tion at  a  glance. 

It  will  be  of  service  to  remember  that  versed-sine 
and  coversed-sine,  and  all  the  functions  of  angles 
of  the  1st  quadi-aut,  are  +  ;  but  tliat  of  the  other 
functions  than  the  versed-sine  and  coversed-sine,  of 


Fig.  6. 


angles  terminating  in  the  other  quadrants,  but  Uco  are  -f-  in  each  quadrant 


*  This  is  a  convenient  elliptical  form  for 
iBt  quadrant,"  etc. 


an  angle  whose  measuring;  arc  terminates  in  the 


10 


PLANE  TRIGONOMETRY. 


ScH.  2. — The  signs  of  functions  of  angles  greater  than  360*  are  readily  deter- 
mined by  observing  in  what  quadrant  the  measuring  arc  terminates.  Thus, 
Bin  570°  is  — ,  since  an  arc  of  570°  terminates  in  the  3d  quadrant.  In  any  given 
case,  the  sign  of  the  function  is  the  same  as  the  sign  of  the  same  function  of  tho 
remainder  left  after  dividing  the  arc  by  360%  or  2;r.  Thus  tan  1180°  is  the  same 
as  tan  100° ;  i.  e.,  it  is  — .  The  same  may  also  be  said  of  the  value  of  the  func- 
tion. 


LDHTENG  YALUES  OF  THE  TRIGONOMETRICAL  FUSCTIOXS. 

83.  Proiy.—Sin  0°  =  if  0,  sin  90°  =  1,  sin  180°  =  ±0, 
«i?i  270°  =  —1,  sin  360°=  tO,  and  the  limits  within  which  the 
sines  of  all  angles  are  comprised,  are +  1  and  —1. 

Deit,— Let  the  point  P  be  supposed  to  start  from  A  and  move  around  the 
circumference  from  right  to  left,  and  let  x  represent  the  angle  (or  arc)  generated. 

Now,  when  P  is  at  A,  a;  =  0,  and  PD  =  0. 

Moreover,  if  we  consider  the  sine  P'"  D'" 

as  reaching  its  limit,  by  the  moving  of  P'" 

from  some  point  in  the  fourth  quadrant  to 

the  origin,  the  sine  of  0°  becomes  —  0,  since 

what  is  true  of  a  taryinrj  quantity  all  the  way 

as  it  appi'oaches  its  limit,  is  assumed  true  at 

tJie  limit.    But,  if  the  sine  reaches  its  limit  by 

the  passage  of  the  point  P  from  some  point 

in  the  first  quadrant  to  the  origin,  the  sine 

of  0°  is  to  be  considered  as  + ,  since  the 

function  is   +   as  it  approaches  its  limit 

.-.  sin  0°  =  T  0.*    As  x  increases  from  0° 

to  90°  (/.  e.,  as  P  passes  from  A  to  a),  PD  is 

+  and  increases  from  0  to  1  (+  0  to  +  1): 

.*.  sine  90°  =  1.    As  a;  continues  to  increase 

from  90°,  sin  x  diminishes  and  becomes  0  at 

Pig.  7.  180°.    To  determine  the  sign  of  sin  180°,  we 

notice  that  it  is  +  M'hen  the  point  P  approaches  this  limit  from  the  second 

quadrant,  and  —  when  it  approaches  it  from  the  third  quadrant, .'.  sin  180°  =  ±  0. 

Conceiving  x  to  go  on  increasing  from  180°,  sin  a;  appears  below  AB  aud  is  — ,  and 

beginning  at  —  0  diminishes  (a  numerical  increase  of  a  negative  quantity  being 

considered  an  absolute  decrease)  till  at  270°  it  becomes  —  1.    .".   sin  270°  =  —  1. 

As  x  passes  from  270°  to  360°,  sin  x  increases  (see  above)  from  —  1  to  0.    The  sign 

of  this  limit  is  ambiguous,  as  appears  by  regarding  tlie  limit  as  reached  by  the 

angle  passing  from  something  less  than  360'  to  360°,  whence  we  have  — ,»nd  also 

as  reached  by  the  angle  passing  hack  from  something  greater  than  360°  to  360°, 

whence  the  sign  is  + .    .-.  sin  360°  =  T  0.    Finally,  as  it  is  evident  that  these 

values  would  recur  in  the  same  order  as  the  point  P  passed  around  again,  the 

above  comprise  all  real  values  of  sines  of  angles.    Q.  e.  d. 

*  It  has  been  customary  to  disregard  the  sign  of  0.  in  such  a  series  as  —  m  .  .  —  3,  —  2, 
—1,  T  0,  +  1,  +  2,  +  3,  ....  +  m,  whereas  it  should  be  regarded  as  ambiguonfl.  as  appears 
above. 


DEFINITIONS  AND  FUNDAMENTAL  RELATIONS.  11 

34.  JProjy.—Cos  0°  =  1,  cos  90°  =  db  0,  cos  180°  =  -  1,  cos  270° 
=  ^0,  cos  360°  =  1,  and  the  cosines  of  all  angles  are  comprised 
hetween  +  1  and  —  1. 

Dem. — Using  the  same  figure  and  the  same  conception  as  in  the  last  demon- 
stration, it  is  evident  that  as  P  approaches  A,  from  either  direction,  that  is  as  « 
approaches  0°,  the  cosine  OD  approaches  to  equality  with  the  radius  and  reaches 
it  at  a;  =  0,  being  +  in  either  case.  .*.  cos  0  =  1.  As  a;  approaches  90°  from  a 
less  value,  i.  e.,  from  the  first  quadrant,  cos  a;  is  +,  and  approaching  +  0  ;  but  as 
X  approaches  90°  from  some  greater  value,  i.  e.,  from  the  second  quadrant,  cos  a;  is 
—  and  approaching  —  0.  .*.  cos  90°  =:  ±  0.  In  like  manner  \ve  obseiTC  that  as 
X  increases  from  90°  to  180°,  cos  x  decreases  (see  Dem.  of  33)  from  0  to  —  1,  which 
it  reaches  at  180°,  and  this,  whether  the  point  is  reached  in  one  way  or  the 
other.  .-.  cos  180°  =  -  1.  Again,  cos  270°  =  T  0  ;  since  it  is  —  0  if  aj  passes  to 
270°  from  some  value  less  than  270°,  and  +  0,  if  it  passes  from  some  value 
greater  than  270.  While  x  passes  through  the  fourth  quadrant,  cos  x  passes 
from  0  to  +  1,  reaching  the  latter  value  when  x  =  360°.  .-.  cos  360°  =  1.  Finally, 
as  it  is  evident  that  the  above  values  would  recur  in  the  same  order  as  the  gen- 
erating point  passed  around  again,  this  discussion  comprises  all  real  values  of  a 
cosine. 


3S.  JPro2).—Tan  0°  =  zp  0,  tan  90°  =  ±  oo,  tan  180°  =  ^  0, 
tan  270°  =  ±  go  ,  tan  360°  =  =f  0,  and  the  tangents  of  all  angles  are 
comprised  between  -t-co  and—a^. 

cosO  1  cos  90         ±  0 

TanlSO-  =  ":  =  ^=  ^  0.  Tan 270°  =  ^-1^  =  ::ii  =  ±  ».  Tau360- 
cos  180         —1  cos2<0        TO 

=■ ^jTTT^  =  -—  =  T  0.    From  these  results  it  appears  that  the  tangent  may 

COS  OOU  J. 

Lave  any  value  whatever  from  +00  to  —  c»  ;  and  as  subsequent  revolutions  of 

the  generatrix  would  evidently  only  repeat  these  values,  these  are  all  the  real 

values  of  a  tangent.    (Moreover,  all  real  values  are  comprised  between  +  oo 

and  —  00).  Q.  E.  D. 

It  is  easy  to  observe  the  direction  of  the  change  in  the  tangent  (whether  it  is 

sin 
increasing  or  decreasing)  by  observing  the  fraction  — ;.    As  the  arc  increases  in 

the  first  quadrant,  the  sine  increases  and  the  cosine  decreases,  for  both  of  which 
reasons  the  tangent  increases,  and  hence  changes  more  rapidly  than  either  sine 
or  cosine.  The  student  should  obseiTC  the  change  in  each  of  the  four  quad- 
rants in  the  same  manner. 

ScH.— These  values  of  the  tangent  are  illustrated  by  Fig.  7.  Thus  AT,  which 
is  +,  becomes  +  0  when  P  returns  to  A,  or  a;  =  0.  Also  AT',  which  is  — , 
and  may  be  considered  as  the  tangent  of  AP'",  a  negative  arc,  becomes  —  0, 
whenAP"'  =  0.  Again,  if  AP  passes  to  Art,  AT  passes  to  +  x  .  But,  if  P' 
passes  hack  to  a,  so  that  AaP'  p;isses  to  Aa,  or  90°,  AT'  passes  to  —  oo.  .*.  We 
see  that  tan  0  may  be  considered  =F  0,  and  tan  90°  =  ±  00 .  In  like  manner  thf 
other  limits  are  illustrated. 


12  PLANE  TEIGONOMETRY. 

SO.  Prop.— Cot  0°  =  ip  o),  co/J  90°  =  =t  0,  cot  180°  =  :?:«>, 
cot  270°  =  ±  0,  co^  3G0°  =  =p  co,  and  the  limits  of  the  cotangent  are 
+  00 ,  and  —  oo . 

Dem.— These  values  are  the  reciprocals  of  the  coiTesponding  values  of  the  tan- 
cos 
gent ;  or  they  may  be  deduced  from  the  relation  cot  =  -r-,  in  a  manner 

altogether  analogous  to  those  of  the  tangent.  Fig.  7  also  affords  geometrical 
illustraticns  of  them.  The  student  should  not  fail  to  deduce  and  illustrate 
them,  and  also  to  observe  the  law  of  change. 


37.  JProp.—SecO°  =  1,  sec  90°  =  ±  od,  5ecl80°  =  -  1,  sec270'* 
=  ip  CO,  sec  360°  =  1,  a7id  all  real  values  of  this  function  are  com- 
prised hetiueen  1  and  ±  oo ,  and  —  1  and  =f  oo  . 

Dem. — These  values  are  the  reciprocals  of  the  corresponding  values  of  the 
cosine.  The  student  should  obtain  them  from  the  cosine,  and  illustrate  them 
from  Fig.  7,  observing  the  law  of  change.  Thus  beginning  at  OA  =  1,  which  is 
the  secant  of  0°,  the  secant  increases  till  x  =  90°,  when  the  secant  OT  becomes 
+  oc,  if  we  consider  this  limit  as  reached  thus ;  but  —  oo ,  if  we  consider 
such  an  arc  as  AaP',  whose  secant  is  —  OT',  which  becomes  —  oo,  when  the 
point  P'  passes  back  to  a. 

Sen. — The  series  which  represent  the  values  of  secants  are,  for  the  first  and 
second  quadrants,  +  1,  +  2,  +  3,  +  .  .  .  .  +  tw,  ±  oo ,  -  w,  —  .  .  .  .  —  3,  —  2, 
—  1;  for  the  third  and  fourth  quadrants,  —  1,  —  2,  -  3,  —  ....—  w,  T  oo, 
+  771,  +  ....  +  3,  +  2,  +  1,  understanding  the  numbei*s  in  these  series  as  rep- 
resenting a  few  terms  of  series  which  have  an  infinite  number  of  terms  of  all 
values  exten(ung,  in  the  first  case,  from  +  1  to  ±  oo ,  and  thence  to  —  1.  It  will 
be  a  good  exercise  for  the  pupil  to  write  the  series  representing  the  values  of 
each  of  the  ti'igonometrical  functions.  Thus,  for  the  sine,  we  have  T  0,  +  i,  +  i, 
+  a^  +  1^  4-  a^  +  ^,  +  ^,  ±  0,  for  the  first  and  second  quadrants,  understand- 
mg  that  all  values  intermediate  between  those  represented  are  included.  For 
tlie  second  and  third  quadiants,  we  have,  +  1,  +  f ,  +  i,  +  i,  ±  0,  -  i,  —  i. 


38.  JPro2)'—Oosec  0°  =  qr  oo ,  cosec  90°  =  1,  cosec  180°  =  =t  oo  , 
cosec  270°  =  —  1,  cosec  3G0°  =  :^  oo ,  and  all  the  real  values  of  this 
function  are  comprised  letween  1  and  ±  oo ,  and  —  1  and  z^  cf^, 

Dem.— Let  the  student  demonstrate  and  illustrate  as  in  the  preceding  article. 
Do  not  neglect  to  go  through  the  whole  in  detail;  it  is  an  important  and 
excellent  exercise. 


39.  Prop.—  Versin  0°  =  0,  versin  90°  =  1,  versin  180°  =  2,  ver- 
sin  270°  =  1,  versin  360°  =  0,  a7id  the  real  limits  of  the  function  aro 
0  and  2. 


DEFINITIONS   AND   FUNDAMENTAL  RELATIONS. 


13 


Dem. — The  student  will  readily  deduce  these  results  from  the  relatiou  versa 
:  1  —  cos  X.    Thus  when  a;  =  0,  vers  0  =  1—  cos  0  =  1  —  1  =0,  etc. 


4z0,  Prop* — Covers  0°  =  1,  covers  90°  =  0,  covers  180°  =  1,  covers 
270°  =  2,  covers  360°  =  1,  and  the  limits  of  the  real  values  of  this 
fmction  are  0  and  2. 

Dem. — The  student  should  be  able  to  give  it 


4:1»  General  Scholium. — It  is  important  to  observe  that  in  the  case  ol 
each  of  the  above  functions  it  cJianges  its  sign  by  passing  through  0  <??•  oo .  In  fact, 
it  is  assumed,  in  mathematics,  that  a  varying  quantity  which  passes  from  +  to 
— ,  does  so  by  passing  through  0  or  oo.  The  converse,  however,  is  by  no  means 
ti'ue ;  viz.,  that  whenever  a  varjing  quantity  passes  through  0  or  oo ,  its  sign 
necessarily  changes.* 


*  The  Co-ordinate  Ge- 
ometry aflbrds  elegant  il- 
lustrations of  the  tlieory 
of  the  change  in  value  and 
sign  of  these  functions. 
(See  Gen.  Geora.,  23, 
etc.)  The  annexed  Fig- 
ure represents  a  curve, 
(or  as  the  student  may 
be  disposed  to  considei 
it,  a  series  of  curves), 
constructed  as  follows  : 
On  the  indefinite  line 
AE»  a  circumference  is 
developed  (as  it  were 
straightened  out),  the 
origin  being  at  A,  a'ld 
AE  being  the  length  o! 
the  circumference.  The 
:urves  mn,  m'n',  r)i"n" 
ire  drawn  by  erecting 
at  every  few  degrees 
from  Ai  a  line  equal 
to  the  tangent  of  the 
same  number  of  de- 
grees, above  the  line 
AE  when  the  tangent  is 
+  ,and  below  AE  when 
the  tangent  is  — .    Thus 

Aa  =  45°  in  length,  and  ab  =  tan  45°  ;  Ac  =  135°,  and  cd  =  tan  135°.  Such  lines  as  ab,  cd,  etc., 
are  called  ordinates  of  the  curve.  The  law  of  change  in  these  ordinates  is  manifestly  the  same 
as  the  law  of  change  in  tangents.  We  see  that  as  we  pass  from  0°  to  90°,  the  ordinate  (tan- 
gent) passes  from  0  to  +  oo.  J»  90°  the  ordinate  (tangent)  in  both  ■¥  and  — ,  i.  «.,  ±  oo.  So  also  at 
270°,  and  at  other  similar  points.  A  similar  device  illustrates  the  changes  in  the  other  trigono- 
metrical functions.  Some  may  see  the  propriety  of  distinguishing  oo  as  -h  and  — ,  who,  never- 
theless, do  not  see  why  it  is  necessary  to  make  the  same  distinction  in  the  case  of  0.  But  a 
moment's  reflection  will  show  that  one  distinction  involves  the  other,  since  oj  and  0  are  muta* 
ally  reciprocals  of  each  other. 


Fig.  8. 


u 


PLANE  TRIGONOMETRY. 


42.  ScH.    The  results  of  the  preceding  discussion  of  the  signs  and  limits  of 
Uie  tiigonometrical  functions,  24  to  41,  are  exhibited  in  the  annexed 


-'1^  the  Limits. 
Between      " 

At  the  Limits. 
Between      " 

W3 

At  the  Limits. 
Between      " 

1 

o 
+ 
3     + 

+ 

tH 

+ 

2     + 

o 
+ 

+ 
2     + 

+ 

+ 
2     + 

+ 

S           1         - 

£           + 

2              3     + 

+ 

2     + 

•1— 1 

+ 

1-' 
+ 

2     + 

+ 

o 

+ 
2     + 

+ 

H 

§ 

+ 
8 

14- 

8 
S     + 

+ 

7 

-    1 

8 

8 
2      i 

i-H 
i 

8 

s    + 

T-t 

+ 

1 

2   1 

8 

8 
H- 
2      1 

1 

+ 

3     + 
8 

H- 

o 
■H 

2     + 
8 

8 
H- 

2      1 

o 

-H 

o 

2     + 
8    ■ 

8 

H- 

K   i, 

o                     8 

H-                         -H 

8 
+1 

2     + 

o 
H- 

H- 

3      1 

8 

H 

o 
-H 
5     + 

+ 

7 

■2      1 

o 

-H 

o 
H- 

2      1 
1 

+ 

2     + 

o 
4- 

+ 

2     + 

o 
H- 

o 
-H 
S       + 

+ 

1 

2      1 

o 

o 
K- 

2      1 

1— < 

1 

ii 

s 

b 

s 

^1- 

s 

DEFINITIONS  AND    FUNDAMENTAL   RELATIONS.  15 


naGOXOMETRICAL  FUNCTIONS  OF  NEGATIVE  ARCS  (OR  ANGLES). 

43,  JProp, — Olianging  tlie  sign  of  an  arc  (or  angle)  changes  tin 
sign  of  its  sine,  and  consequently  of  its  cosecant;  i.e.,  sin  (— a;)  = 
—  sin  a;,*  and  cosec  (—  a:)  =  —  cosec  x. 

Dem.— 1st,  If  the  angle  (or  arc)  is  numerically  Uss  than  90',  and  + ,  it  ends  in 
tlie  first  quadrant,  and  hence  its  sine  S%  +  ;  but,  if  the  angle  (or  arc)  is  numeri- 
cally less  than  90°  and  — ,  it  ends  in  the  fourth  quadrant,  and  lience  its  sme  is  — . 
That  is,  X  being  numerically  less  than  90°,  sin  (—  a;)  =  —  sin  x.  2d,  If  a;  is 
between  90"  and  180°  in  numerical  value,  the  arc  ends  in  the  second  quadrant 
when  a;  is  + ,  and  hence  its  sine  is  +  ;  but  when  a;  is  — ,  the  arc  ends  in  the  third 
quadrant,  whence  its  sine  is  — .  .'.  In  tliis  case,  also,  sin  (—  a;)  =  —  sin  x,  3d, 
If  X  is  between  180°  and  270°  in  numerical  value  and  + ,  the  arc  ends  in  the 
tliird  quadrant,  whence  sin  a;  is  — ,  (—  sin  x) ;  but  if  the  arc  is  — ,  it  ends  in  the 
second  quadrant,  whence  its  sign  is  +,  [+  sin(—  x)\.  :.  In  tliis  case  —  sinaj 
=  sin  (—  a;),  or  sin  a;  =:  —  sin  (—  a;).  4th,  In  like  manner  the  student  will  observe 
that  —  sin  a;  =  sin  (—  a;),  or  sin  a;  =  —  sin  (—  x\  when  x  is  between  270°  and  3G0°. 
Moreover,  since  this  order  will  recur  as  we  pass  around  again,  we  learn  that 
in  any  case  the  sign  of  the  sme  of  a  negative  angle  is  the  opposite  of  that  of  an 

equal  positive  angle.     Finally,  cosec  (—  a:)  =  — — ; =  — -. —  = =  — 

•"  ^      ''      sm  (—  a;)      —  sm  a;  sin  x 

cosec  X.     Q.  E.  D. 


44.  Trop.—Clianging  the  sign  of  an  angle  {or  arc)  does  not 
change  the  sign  of  its  cosine  or  secant ;  i.  e.,  cos  {—  x)  =  cos  a:,  and 
sec  {—  x)  =  sec  a;. 

Dem.— 1st,  When  x  <  90°  and  + ,  the  arc  ends  in  the  first  quadrant,  and  hence 
its  cosine  is  +  ;  so,  also,  if  a;  <  90°  and  — ,  though  the  arc  ends  in  the  fourth 
quadrant,  its  cosine  is  still  + .    .'.  cos  (—  a;)  =  cos  a;.  (The  student  can  supply  the 

other  three  cases.)    Finally,  sec  (—  a;)  = = =  sec  x.    o.  e.  d 

cos  (—a:)      cos  a;  ^ 


45,  Vrop,— Changing  the  sign  of  an  angle  {or  arc)  changes 
the  sign  of  its  tangent,  and  consequently  of  its  cotangent;  i,  e., 
tan  (—  cc)  =  —  tan  x,  and  cot  (—  ic)  =  —  cot  x. 


♦  The  etudent  should  be  careful  to  note  the  exact  meaning  of  this  expression.  It  is  read 
"Bine  minus  x  -  minus  sine  a;,"  and  means  that  the  sine  of  a  negative  angle  (or  arc)  i8  equR 
(numerically)  to  the  sine  of  the  same  positive  angle  (or  arc),  but  has  Jhe  opposite  sign. 


16  PLANE  TRIGONOMETET. 

Dem. — This  is  an  immediate  coDsequence  of  tlie  fact  that  changing  the  sigc 

of  the  angle  (or  arc)  changes  the  sign  of  its  sine,  but  not  of  its  cosine.    Thus, 

sin  (—a-)      —sin  a:  sin  a;  ^  .,  4./       x  1 

tan (-  x)  =  . — i^ — '-  = = =  -  tana!.  Also,  cot  (-  x)  =  - — ; 

cos  (—a-)        cos  a;  cos  a;  taE(— ar) 

1  1  ,  ^ ,       .       cos  (—a-)        cos  X  cos  x 

= ^ — = = — =  — cota;;    or  cot(— a:)  = -v— ^ — f= -. —  = ■. — 

—  tan  aj  tana;  sm(— a;)       —  sma;  sin  a 

=  —  cot  a;. 

Sen. — The  proper  sign  of  a  function  of  a  negative  angle  can  always  be  ascer- 
tained by  observing  in  what  quadrant  the  measuring  arc  cuds,  in  a  maunei 
altogether  similar  to  that  in  which  the  signs  of  the  functions  of  positive  angles 
are  deteniiined. 


CIRCULAR  FUISCTIONS. 

46.  Circular  Functions  are  angles  (or  arcs)  expressed  as 
functions  of  sines,  cosines,  tangents,  or  other  trigonometrical  lines. 

Ill's. — In  the  expression  sin  x,  we  designate  a  sine,  i.  e.,  a  right  line,  merely 
using  the  x  to  tell  wlmt  sine,  as  the  sine  of  20°,  of  135°,  etc.  But  we  often  wish 
to  speak  of  an  angle  (or  arc)  which  has  a  particular  sine,  tangent,  or  other  trigo- 
nometrical line.  Thus,  we  say,  "  the  angle  (or  arc)  whose  sine  is  ^,'*  "  the  angle 
(or  arc)  whose  tangent  is  3,"  etc.  In  this  form  of  expression,  it  is  evidently  the 
angle  (or  arc)  which  is  the  thing  mamly  thought  of;  and  it  is  conceived  as  de- 
pendent upon  its  trigonometrical  line. 

47.  Notation. — The  circular  functions  are  written  sin"'?/,  cob~^x, 
tan-'z,  etc. ;  and  are  read  "the  angle  (or  arc)  whose  sine  is  y"  "the 
angle  whose  tangent  is  2;,"  etc. 

Ill's. — The  expressions  x  =  s'm-'^y,  and  y  =  sin  x,  are  ultimately  equivalent, 
since  the  first  is  "  a;  =  angle  whose  sine  is  y,"  and  the  second, "  y  =  the  sine  of  a-." 
The  only  difference  is,  that  in  the  first  form  the  angle  (x)  is  the  thing  thought  of, 
and  the  sine  {y)  is  used  merely  to  tell  wJiat  angle ;  but  in  the  second  form,  the  stTis 
iy)  is  the  prominent  thing,  and  the  angle  ix)  is  used  simply  to  tell  wMt  sine.  This 
mutual  relation  has  caused  the  circular  functions  to  be  called  also  Inverse  Func- 
tions. 

ScH. — This  notation  is  rather  an  unfortunate  one,  inasmuch  as  it  is  the  same 
as  has  been  already  adopted  in  the  theory  of  exponents.  The  student  will  how- 
ever obseiTe  that  the  signification  in  this  instance  is  altogether  different  from 
the  former.    Thus,  since  we  write  "  the  square  of  sme  a*,"  sin'' a;;  according  to  the 

theory  of  exponents,  sin-'rc  would  be  -7-—-.   So  also  sin-^a;  should  mean  -, 

sm-'a;  sm  x 

Now,  the  former  of  these  expressions  would  actually  signify  as  indicated  (though 

it  were  better  to  write  it  (sin  a-)-'),  while  the  latter  does  not  mean  at  all  what 

the  theory  of  exponents  would  make  it.    Unfortunate  as  the  notation  is,  it  is 

probably  best  to  retain  it.    It,  doubtless,   was  suggested  thus :   If  we  have 

y  =  rt'a;,  we  may  write  x  =  a-'y,  so  also  y  —  ax,  may  I: e  written  x  =  a-^y. 

This  affords  a  parallelism  in  form,  but  not  in  signification. 


DEFINITIONS   AND   FUNDAMENTAL    EELATIONS. 


17 


EXERCISES. 

1.  Whiit  IS  the  complement  of  150°  21'  13".5  ?     Wheat  the  sup- 
plement?     Give  the   complements   and  also   the   supplements   of 

125°  15',   283°  21'  11",  36°  05'  02",   and  89°  00'  12". 

If 

9 


2.  How  many  degrees  in  the  angle  (or  arc)  -o  ?     In  ^-rr  ?      In   the 

arc  2*?    Ini*?     InlJ-r? 

3.  How  many  times  is  the  radius  contained  in  108°  ?  IIow  many 
times  in  2'rt  ?     How  many  in  460°  ?     How  many  in  210°  ? 

4.  Radius  being  taken  as  the  measure  of  the  arc,  by  what  are  45° 
represented?  By  what  90°?  By  what  180°?  By  what  225°? 
How  many  degrees  does  1  represent,  radius  being  the  measure  ? 

5.  Find  the  length  of  a  degree  of  the  meridian  upon  a  globe  of  18 
inches  diameter. 

6.  Express  12°  22'  13"  11'"  05'^  in  °, ',  ",  and  decimals  of  a  second. 
So  also  express  53°.51  in  °,  ',  ",  etc. 

7.  How  many  degrees,  minutes,  and  seconds  in  an  arc  equal  to 
twice  radius?  three  times  radius?  Show  that  27°  is  equivalent  to 
^^if.  That  10°  to  radius  10  ft.  =  1.75  ft.  Wliat  radius  gives  1°  = 
1  inch  ? 

8.  Draw  any  angle,  and  construct  its  sine,  cosine,  tangent,  or  any 
other  trigonometrical  function,  and  then  determine  as  ne'arly  as  prac- 
ticable the  nivmerical  values  of  the  function  by  actual  measurements 

Solution. — Given  the  angle  MON,  to  find  the  numerical 
value  of  the  tangent  as  near  as  practicable  by  measurement. 
Taking  any  convenient  unit,  as  OA,  for  a  radius,  and  striking 
the  arc  A  a,  draw  AT  tangent  ♦o  a/K  at  A.  Now  apply  OA  to 
AT  and  find  their  ratio  (Part  I.,  30).  In  this  case  AT  =  1^ 
approximately, .'.  tan  MON  =  1^. 

[Note. — The  student  should  practice  upon  such  exam- 
ples, finding  the  values  of  all  the  functions  until  the  process, 
and  the  meaning  of  the  numerical  value  of  any  function 
of  an  angle,  are  clearly  seen.] 

^  9.  Construct  an  angle  whose  sine  is  f ,  i.  e.,    ' 

Solution. — Let  0  be  the  required  angular  point, 
and  OA  one  side  of  tlie  angle.  Lay  off  from  0  on 
OA  3  measures  of  any  convenient  length,  making 
Ort.  Using  Oa  as  a  radius,  describe  the  indefinite 
arc  aM.  Erect  OC  perpendicular  to  OA  and  take 
OC  =  ^  of  Oa.  Through  C  draw  CP  parallel  to 
OA.  Finally,  draw  OB  through  P.  AOB  is  the 
angle  required,  since  Oa  being  1,  the  sine  of 
AOB,  PD,  is  i     AOB  =:  siu-H. 

Fig.  l(k 


Fig.  9. 


5  in" 


18  PLANE  TRIGONOMETRY. 

10.  Construct  an  angle  whose  cosine  is  §.  That  is,  Cjnstnict 
cos~'|. 

SuG.— The  constiiiction  is  tbe  same  as  in  Jlie  last  example,  except  th.at  instead 
of  OC  being  drawn  to  limit  the  arc,  a  perpendicular  is  erected  to  Oa  at  i  (the 
second  point  of  division)  from  0,  and  the  point  P  located  where  this  perpendicular 
intersects  the  arc. 

11.  Construct  an  angle  whose  tangent  is  2.  That  is,  construct 
tan-'2. 

SuG.— To  construct  this  at  0  on  the  line  OA,  Fig.  10,  take  any  convenient  ra- 
dius, Oa,  and  strike  the  indefinite  arc.  Then  erect  at  a  a  tangent,  and  make  it 
equal  to  twice  the  radius  used.  Through  the  extremity  of  this  tangent  and  O 
draw  a  line,  and  the  angle  between  this  and  OA  is  tan-'2. 

12.  Construct  sec~'2 ;  cot~'3;  cosec~^l^;  an  obtuse  angle  sin~'^; 
tan-'(-3). 

SuG. — To  construct  sin-^^,  see  Fig.  7.  Let  OA  be  one  side  of  the  angle,  and 
0  the  vertex.  With  any  convenient  radius  draw  the  semicircumference  AaB, 
and  draw  the  perpendicular  Oa.  Bisect  this  perpendicular,  and  through  the 
point  of  bisection  draw  a  parallel  to  AB,  intersecting  the  arc  in  the  2d  quad- 
rant. Through  this  intersection  dmw  a  line, as  OP'.  Then  AOP'  (assuming  the 
construction  as  specified,  and  not  as  in  the  figure)  =  sin-^i. 

13.  Construct  the  following:  tan-'l  ;  tan-'(  — 1);  ta,n-'J ; 
tan-'(-2);  tan-'(-i) ;  cos-'(-i);  sec-'(-2);  cosec-'(-3) ; 
versin~4  ?  versin~'lj. 

14.  From  the  fundamental  relations  (21)  deduce  the  following : 

sni.^=\/l  — cos'.'?:;  cosa;=  Vl— sin^a:;  tan.'ccota;=l;  tan2;cosa;=sin.T; 

sin  a:      .  cosa:       .  ^       -pn^vr—  ^ 

cos3rc  =  - :sma;= — : — :sin.T=  ,cos.r—  -.. 

tan.T  cot.r  V 1-1- cot  .'c  vl  +  tana;' 

sec.-c 

tan.T  =  sma;  seca;;    cota:  =  cos  a;  cosec.^;  tan  re  = ;  sm.'c  = 

'  cosec.^' 


cos  a:                           secT— 1      cosecr— cot.T        ,  \/l — sin-o; 

"    vers  .r  =  = ;  cota;  = 


Vcosec'a;— 1  sec.-c  cosec  .^•  s'mx 

15.  Given  tana;=:|,  to  find  the  other  trigonometrical  functions 
of  X. 

Results :   Sec  a;  =  f ;   cos  .t  =  | ;   sin  a;  =  | ;   cosec  a:  =  | ;   cot  a 
=  f ;  vers  x  =  \\  covers  x  =  \. 

16.  Given   sin  a;  =  f ,  to  find  the  other  trigonometrical  functions 
of  X. 

17.  Given   sec  a:  =  2,  to  find  the  other  trigonometrical  functions 
of  X. 

18.  Given  tana:  =  —  1,  to  find  the  other  trigonometrical  functionfa 

Df  X. 


DEFINITIONS   AND   FUNDAMENTAL   EELATI0N8.  19 

Results. —Cot  X  =  —  1;  sec  re  =  rp  V^;  cosec  x  =  dzA/T;  sin  x 
=  ±  J  V2~;    cosx  =  zpl^/2;  versin  x  =  1  zh  ^y^;  coversin  a;  = 

[Note. — Observe  closely  the  signs  of  the  functions  in  Ex.  18.] 

19.  In  the  preceding  examples  the  constructions  required  have 
been  limited  to  angles  less  than  180°,  but  it  is  evident  that  an  infi- 
nite number  of  angles  (or  arcs)  according  to  the  more  comprehensive 
trigonometrical  view,  correspond  to  the  same  function.  What  angles 
or  arcs  have  their  tangents  each  1  ?  What,  each  —  1  ?  Construct  an 
angle  between  130°  and  270°,  whose  sine  is  —J.  What  other  angle 
less  than  360°  has  the  same  sine  ? 

20.  Having  given  a  sine,  how  many  angles  less  than  180°  corre- 
spond to  it  ?  Construct  the  angle  or  angles  less  than  180°  whose  sine 
is  |.  How  many  angles  less  than  180°  have  the  same  cosine  ?  tan- 
gent ?  cotangent  ?  (In  each  of  the  last  three  cases  07ih/  one.)  Con- 
struct ^  =  cos-' J;  y  =  cos-' (—J).  How  are  these  angles  related 
to  each  other?  Construct  ?^  =  tan-'3;  y  =  tan-'(  — 3).  How  are 
these  angles  related  to  each  other?  (Restrict  the  constructions  and 
questions  in  this  example  to  angles  less  than  180°.) 

21.  Given   y   =   sin-\T,   show    that   cosy    =   yi  — re';   tany  = 
X  1  1 


22.  Wliat  are  trigonometrical  functions  of  450°  ?    Of  1350°  ?    Of 
900°  ? 

23.  Show  that  the  following  are  true  for  all  integral  values  of  n, 
including  0 :   sin  4?i^  =  0 ;   sin  (4;t  +1)^=1;  gin  (4n  +  2)  J^  =  0 ; 

Bin(4?i  +  3)---l;cos47J^  =  l;  cos  (4?i +1)  ^  =  0;  cos(4?i4-2)^ 

=  -  1 ;  cos  {4.n  +  3)^  =  0;  tan  2?i^  =  0 ;  tan  (2?i  +  1)  ^  =ro . 

24.  What  are  the  signs  of  the  several  trigonometrical  functions 
of -110°?     Of -35°?     Of -500°?     Of -2000? 

25.  Prove  that  sin  30°=  i,    cos  30°  =  ^^3,  tan  30°=  ^a/S,  and 
cot  30°=  V3,  sec  30°=  I V3,  and  cosec  30°=  2. 

SuG.— Observe  that  the  chord  of  60°  =  1,  and  that  the  sine  of  30°=  ^  tht 
chord  of  60°.    Make  the  figure. 


20 


PLANE   TEIGOXOMETKY. 


/  26.  Prove   that  siu  45°  =  iv^,  cos  -45"=  ^V^,  tau  45"=  1,  cot  45' 
/=  1,  sec  45°=  V^,  and  cosec  45°=  v^. 

! 

ScG. — Observe  that  sin  45*  =  cos  45° ;  Lence  sin'  45'^  +  cos'  45°  =  1,  becomes 
2sm'  45°  =  1. 

Sen. — The  vulues  obtained  in  the  last  two  examples  should  be  retained  in  the 
Diemory,  as  they  are  of  frequent  use.  The  functions  of  30°  and  of  45°  are 
*ilway3  assumed  to  be  known  in  any  trigonometrical  operation. 


SECTIOX  11. 


RELATIO>'S    BEr\>T:E>    THE    TRIGOXOXETRICiX  FU>CTIO>S  OF 
DrFFERE>T  1>GLES  (OR  ARCS). 

(a)    FUKCTIOXS   OF  THE   SUM   OR   DIFFERENCE   OF   ANGLES 
(OR   arcs). 

4,8,  I^rop, — Tlie  sine  of  the  sum  of  tivo  angles  {or  arcs)  is  equal 
to  the  sine  of  the  first  into  the  cosine  of  the  second,  phis  the  cosine  of 
the  first  into  the  sine  of  the  second.  Thus  letting  x  and  y  represent 
any  two  angles  (or  arcs), 

sin  {x  +  y)  =  sin  a;  cos  ^  +  cos  x  sin  y. 

Dem.— Let  AOB  and  BOC  be  the  two  angles 
represented  respectively  by  x  and  y.  Draw  the 
measuring  arc  aP\  and  the  sines  PD,  aud  P'E  of 
the  angles.  AOC  =  AOB  +  BOC,  is  the  sum  of 
the  two  angles.  Draw  P'D',  the  sine  of  the  sum 
of  the  two  angles.  Then  PD  =  sin  x,  P'E  =  sin  y, 
OD  =  cos  a:,  OE  =  cos  y,  P'D'  =  sin  (.»  +  y),  aud 
CD'  =  cos{x-\-y). 

Now  sin(j:  +  y)  =  P'D'  =  EF  +  P'L.    But  from  the  similar  ti'iangles  EOF, 

EP    _  sin  x 
cos  y         1 


D'  F  D  ^     A 

Fio.  11. 


ana  POD,  we  have  ;|^^  =  ??,  or 


OE        OP 
the  similar  triangles  P'EL  and    POD,  we  have    p,^ 


EF  =  sin  J  cosy.    Also,  from 

P'L         OD    ...     P'L  C0S2 


ODor   P^ 


OP        smy  1 

.-.    PL  =  cos  a;  sin  y.      Substituting  these  values   of    EF  and    P'L,  we  have 
sill  (.?  +  y)  =  sin  a;  cosy  +  cos  a;  siu  y.    q.  e.  d.* 

*  This  demonstration  may  seem  defective,  eince  the  snm  of  the  angles  z  and  y,  as  represented 
in  the  diagram,  is  less  than  90*  ;  nevertheless,  in  the  General  (Analytical)  Geometry,  we  con- 
Ftautly  proceed  in  a  manner  entirely  analogous ;  viz.,  first  produce  the  equation  of  a  locua 
from  some  particular  fi^Tire,  and  then  make  it  general  in  application.  The  demonstrations  in 
cases  in  which  the  sum"of  the  angles  is  greater  than  90=",  etc.,  are  similar,  and  some  of  them 
will  be  given  in  the  Exercises  at  the  close  of  the  section.  It  is  not  thought  best  to  cumber  thtr 
purely  theoretical  part  of  the  subject  with  such  matter. 


FUNCTIONS   OF  THE   SUM   OR   DIFFEEENCE   OF  ANGLES.  21 

Cor.    Si?i{90°  +  x)=cosx.    Sin{lSO°+x)  =  -smx.  Sin{270°  i-x) 
=  —  cosx.     Si7i{SQ0°  -i-  x)  =  sinx. 

Dem.— From  the  proposition  we  have, 

sin  (90°  +  x)  —  sin  90°  cos  x  4-  cos  90°  sin  x  =  cos  x ; 
since  sin  90°  =  1,  and  cos  90°  =  0. 
In  like  manner, 

sin  (180°  +  a:)  =  sin  180°  cos  a;  +  cos  1 80°  sin  a;  =  —  sin  a? ; 
since  sin  180°  =  0,  and  cos  180°  =  —  1. 

[The  student  will  readily  produce  the  other  forms.] 


49.  JProj^. — T/ie  sine  of  the  differencG  iettueen  two  angles  (or  arcs) 
ts  equal  to  the  sine  of  the  first  into  the  cosine  of  the  second,  viinus  the 
cosine  of  the  first  into  the  sine  of  the  second.  Thus,  letting  x  and  y 
represent  the  angles, 

sin  [x  —  y)  =  sin  x  cos  y  —  cos  x  sin  y. 

De-m.— In  sin(:z;  +  V)  =  sin  a;  cosy  +  cos.r  siny,  substituting  —y  for  y,  Ave 
have,  sin  {x  —  y)  —  sin  x  cos  (—  y)  -^  cos  x  sin  (—  y)  =  sin  ^  cos  y  —  cos  a;  sin  y  ; 
since  cos  (—  y)  =  cosy,  and  sin  (—  y)  =  —  siny.    (45,  4:4.) 

Cor.  Sin(^0'' -x)  =  cos  x.  Sin(lSO°-x)  =  sin  x.  Sin{270°-x) 
=  —  cosx.    Sin  (360°  —  x)  =  —  sinx. 

Dem. — This  is  simply  an  application  of  the  proposition,  as  the  preceding 
corollary  is  of  its  proposition.  (The  student  will  make  it.)  Or  we  may-deduce 
the  results  from  the  corollary  under  the  preceding  proposition  by  merely  sub 
stituting  —  X  for  x.  Thus,  sin  (90°  —  x)  =  cos  (—  x)  =  cos  a; ;  sin  (180°—  x)  — 
—  sin  {—  x)  =  —  (—  sin  x)  =  sin  x  ;   etc. 


50.  JProj), — TJie  cosine  of  the  sum  of  two  angles  (or  arcs)  is  equal 
to  the  rectangle^  of  their  cosines,  minus  the  rectangle  of  their  sines. 
Thus,  letting  x  and  y  represent  the  angles, 

cos  (a;  +  y)  =  cos  a;  cos  ?/ —  sin  a;  sin  ?/. 

Dem. — Taking  the  formula  sin  {x  —  y)  =  sin  a;  cosy  —  cos  x  siny,  and  substi- 
tuting 90°—  X  for  X,  we  have,  sin  (90°  —  x  —y)=  sin  (90°  —  x)  cos  y  —  cos  (90' 
-  X)  sin  y.  Now  90°  -  a;  -  y  =  90°  -  {x  +  y) ;  and  sin  [90°  -  (a?  +  y)]  =  the 
sine  of  the  complement  of  {x  -f-  y),  or  cos  {x  +  y).  Also  sin  (90°  —  a;)  =  cos  ar 
and  cos  (90°—  x)  =  sin  x.   .'.  Substituting,  cos  {x  -f-  y)  =  cos  a;  cos  y  —  sin  x  sin  y. 

CoR.  Cos(dO°  +  x)=-si7ix.  Cos(lSO°  +  x)=-cosx.  Cos('270°+z) 
=  sin  X.     Cos(3G0''  +  x)  =  cos  x. 

Dem. — Apply  the  proposition. 


22  PLANE   TKIGONOMETRY. 

51,  JProp. — Tlie  cosine  of  the  dijference  ofUvo  angles  {or  arcs)  is 
equal  to  the  rectangle  of  their  cosines,  j^lns  the  rectangle  of  their  sines. 
Thus,  letting  x  and  y  represent  the  angles, 

cos  {x  —  y)  =  cos  X  cos  y  +  sin  x  sin  y. 

Dem.— In  cos  {x+  y)  =  cos  a;  cos  y  —  sin  x  sin  y,  substituting  —  y  for  y,  wo 
have,  cos  {x  —  y)  =  cos  x  cos  (—  y)  —  sin  a:  sin  (—  y),  or  cos  {x  —  y)  ~z  cos  x  cos  » 
+  sin  a;  sin  y ;  since  cos  (—  y)  =  cos  y,  and  sin  (—  y)  =  —  sin  y.  [Nctice  that  the 
last  term  becomes  —  sin  x  (—  sin  y),  which  equals  +  sin  x  sin  y.] 

Cor.  a?5(90°-rc)=5i/ia:.  Cos{lSO''-x)=  -  cosx.  Cos(270°-x) 
=  —  si';i  it*.     Cos  (360°  —  ic)  =  cos  a:. 

Dem. — Apply  tlie  proposition. 


32,  ^rop. — Tlie  tangent  of  the  swn  of  ttvo  angles  {or  arcs)  it 
equal  to  the  sum  of  their  tangents,  divided  by  1  mimis  the  rectangh 
of  their  tangents.    Thus,  x  and  y  being  the  angles,  we  have, 

tan  X  +  tan  ii 

tan  {:c  +  y)  =  -^ : — ^ 

^        ^^       1  —  tan.^  tun^ 

sin  (x  +  v)      sin  x  cos  v  +  cos  x  sin  y     ,^.   . ,. 

Dem     Tan  (x  +  v)  =  = ^—- — ■. ^-     Dividmg  numer- 

UEM.     1  an  v^  -h  y)       ^^^  {x  +  y)      cos  xcosy-  sin  x  sin  y 

ator  and  denommator  of  the  last  fraction  by  cos  x  cos  y*  we  have,  tan  {x  +  y)z= 

sin  X       sin  y  sin  x       sin  y 

?;;^  "^  ^  ^     ^^^  coflj/     ^  tana;  +  tan  y,  ^^^^  !!^  =  tanar,  etc.,  and 


sin  a;  sin  y  sin  a;       siny      1  — tan  a;  tan  y  cosa; 

cos  X  cos  y  cos  x     cos  y 

sin  X  sin  y  _  sin  x       siny      ^    j^   p, 
cos  X  cos  y       c^)s  a;      cos  y'   ' 


Cor.     Trt?i(90°+rc)=  -6-0/^2:.  Tan{lSO°  +  x)  =  tanx.  Ta7i{270°-hx) 
=  -  cotx.     ra7i(360°+  x)  =  tan  x. 


*  The  three  following  fiorms  may  readily  be  obtained  by  dividing  reepectively  by  sin  x  sin  y, 

cot  ?/  +  cot  X      ^      ,  ^      1  +  cot  a;  tan  y        , 

sin  X  cos  y,  and  cos  x  sin  y :  viz.,  tan  (x  +  y)  =  '^^^J^—'i:  *^°  fa?  +  y)  =  ^^^^_^^j^y>  and 

tan  (X  +  ?/)  =  ^^"  ^  '^^  ^— — .    Any  one  of  these  may  be  reduced  to  the  one  given  in  the  prop- 
cot  y  —  tanx 

osiiion,  bv  substituting  in  it  —  for  cot,  and  reducing.    Notice  that  the  form  in  the  Prop,  is  ic 
tan 

terms  oJ  the  tangents.    Also  observe  tchy  dividing  by  a  certain  term  gives  a  particular  form, 

ciid  by  which  to  divide  to  get  a  required  form. 


FUNCTIONS  OF  THE  SUM  OR  DIFFERENCE  OF  ANGLES.     23 

Dbm.    Tan  (90°  +  x)  =  'i^i^^51jl5i!  =  _5£i^  =  _  cot  a:.     Tan  (180^  +  x)  =.- 
'      cos  (90   +  «)      —  sinaj  ^  ' 

sin  (180°  -V  X)       —  sin  a;      ^ 

— r-r-- -L  = .  =  tan  X,  etc.,  etc. 

cos  (180    +  a?)      —  cos  aj 


53,  I*rop, — The  tangent  of  the  difference  of  two  angles  {or  arcs) 
is  equal  to  the  difference  of  their  tangents^  divided  by  1  j^lfis  the  rect- 
angle of  their  tangents.    Thus,  x  and  y  being  the  angles, 

,  .         tan  X  —  tan  ii 

tan  (x  —  y)  = : — -  . 

^        '''       1  +  tan  x  tan  y 

rx  m      /  .       sin  (o;  —  y)       sin  x  cos  y  —  cos  x  sin  y 

Demonstkation.     Tan  (a;  —  j^)  =  \ ^  =  "- ,- zJtM  ^ 

cos  [x  —  y)       cos  X  cosy  +  sin  x  sin  y 

Bin  X  cos  y       cos  a;  sin  y  sin  a;  _  sin  y 

cos  a;  cosy       cos  a;  cos  v  cos*       cosy  tana;— tan, y 

-1 . __^  —  ; JL —  — ^_.     Q,  Y^   D^     (gj^e 

COS. a;  cos  y       sin  .csiny       .       sin  a;       sin  y       1  +  tan  a;  tan  y 

cos  ar  Cos  y       cos  a;  cos  y  cos  x       cos  y 

foot-notes  to  preceding  proposition.)    This  proposition  is  also  readily  deduced 

from  tlie  preceding  by  substituting  in  tlie  formula  tan  {x  +  y)=z    ^Q  ^  +    J^"  V  ^ 

1  —  tan  X  tan  y 
—  y  for  y,  and  remembering  tliat  tan  {—  y)  =  —  tan  y. 


Cor.    Tan{^^°-x)=  cotx.  Tan{1^0°-x)=  -tanx.  ^«?^(270°-.^•) 
=  co^.T.     Trt?i  (360°—  x)  =  —  tan  x. 


DKM.     Taa  (90=  -  .)  =  '^^:^,  =  |^  =  cot ...      Tan  (180=  -  ., 

?i^-^=.^^  =  -tau..  Tan(270»-^)=^i5^„;^=:^^2ii 
cos.(180  —  a;)      —cos  a;  cos  (270    —  a-)     —  sin.c 

m      •o/.Ao       X       sin(3G0°— a;)       —  sin.i; 

cot  a?.    Tan  (360°-  x)  =  — )--^ {  = =  -  tan  x. 

'      cos  (800*^— a))         cos  a; 


*  These  and  the  kindred  formula  maybe  produced  by  a  direct  application  of  the  proposition. 
n,u       ^      mn«         ^       tan  90°  +  tan  x  tan  90"  1 

Thus,  tan  (90o  -h  x)  =  --^^^^^^  =  ^:^^^^^  =  ZI^^  =  "  <^^'  ^'    <The  reason  foz 

dropping  tana;  and  1  is  that  they  are  finite  terms  connected  with  infinities,  as  tan  90^  =  oo.  Or, 

tana;  .      tana; 

■ ->.   otherwise,  tan  (90°  +  a;)  =  ;^"»0°-^tanx  ^       "■  ^K^^^  ""  "^  =  _1_     .j^^e   a 

^      1  — tan 90°  tana;  1  1  —tana:'  ^ 

* — t^f:^  —  tan  a; tan  a; 

tan  90°  00 

finite  divided  by  an  infinite  equals  0.    Again,  tan  (180°  +  x)  =  i£!Li^±i^.^_  =  1^E£  .-, 

1  —  tan  180°  tan  x         1 
tana;,  since  tan  180°  =  0,  etc.,  etc. 


24  PLANE   TRIGONOMETRY. 

o4z.  Projy- — Tlie  cotangent  of  the  sum  of  two  angles  {or  arcs)  u 

equal  to  tlie  rectangle  of  their  cotangents  mimis  1,  divided  ly  their 

sum.    Thus,  X  and  y  being  the  angles, 

,  ,  .       cot  X  cot  ?/  —  1 

cot  (x  +  y)  = ^  ,  .  . 

^        ^^        coty  +  cot  a; 

COS.T  cosy  sinx  siny 

cos(^  +  ?/)  _  co&r  cosy  —  sin.r  siny  _  sin.c  siny  sin.r  sin^ 

Sin  (^  +  y)      sinxcosy  4.cos^  smy      sin^cosy  cos^smy 

siiLC  siny  sulc  siny 

cos  X     cos  y 

sin.r      siny  cota:coty— 1  _  i    i         -^   xi 

= = .    Q.  E.  D.    Or  we  may  deduce  it  tlms, 

cosy       cosi;  coty  +  cot  a; 

siny       sin  x 

1-     ' 


1  _  1  —  tan  X  tan y cot  x  coty  _  cot  x  cot  y  —  1 

tan(i;  +  y)  ~    tan  ^  +  tany   "~      1         _1 ""     coty  +  cota; 

cot  J      coty 

Q.    E.   D. 

Cor.    Cot{dO°-h  x)  =  ^  tanx.   Cot{lSO°+  x)  =cotx.  Cot{270°-\-x) 
=  -  tan  X.     Cot  (360°+  x)  =  cot  x. 

DEir.    Divide  cosine  by  sine,  or  take  tlie  reciprocals  of  the  corresponding 

tauircnts     Thus,  cot  (90°  +  «)  =  - — j^r^^ r  = —  =  —  tan  a;,  etc. 

'        ^  '      tan  (90    +  a)       —  cotx 


5o,  ^vop, — Tlie  cotangent  of  the  difference  of  ttuo  angles  (or 
arcs)  is  equal  to  the  rectangle  of  their  cotangents  plus  1,  divided  by 
their  difference.     Thus,  x  and  y  being  the  angles, 

,  .  _  cot  X  cot  y  +  1 

^        ^^  ~    coty  —  cot  a: 

Dem.  Substitute  —  y  for  y,  in  the  preceding  formula ;  or,  divide  cos  [x  —  y) 
by  sin  (.c  —  y)  and  reduce  ;  or,  take  the  reciprocal  of  tan  {x  —  y),  and  substitute 

—  for  tan. 
cot 

Cor.  Cot  (90°  -  x)  =  tanx.  Cot  (180°~a:)  =  -  cotx.  Cot (270°-  x) 
=  tan    X.      Cot(dC)0°—x)  = —cotx. 

Dem.     Same  as  above. 

ScH.  1.  The  farmulm  for  the  secant  and  cosecant  of  the  sum  and  of  the 
difference  of  two  angles  (or  arcs)  are  not  of  sufficient  importance  to  waiTant 
their  introduction  here ;  some  of  them  will  be  given  in  the  exercises,  as  also  the 
extension  of  those  already  given  to  the  case  of  the  sum  of  three  or  more  angles, 
or  arcs. 


FUNCTIONS  OF  THE  SUM  OR  DIFFERENCE  OF  ANGLES. 


25 


Ben.  2.     The  results  reached  in  this  discussion  are  so  important  that  we  will 
rollect  them  into  a 


(A)  sin  {x  +  y) 

(B)  sin  {x  -  y)  -. 

(C)  C(w(.t!  +  y)  - 

(D)  cos(.^  -y)- 

(E)  tan  {x  +  u): 

(F)  tan  (^-2^): 

(G)  cot  {x  +  y) 
(H)  cot  {X  -  y) 


TABLE 

:  sin  X  cos  y  +  cos  x  sin  y. 
sin  X  cos  y  —  cos  x  sin  y. 
cos  X  cos  y  —  smx  sin  y. 
cos- a;  cosy  +  sln.c  siny. 

tan  X  +  tan  y 
1— tanaj  ismy 

tana;  —  tan  j/ 
1  +  tana;  tan  y* 
cot  a;  coty  —  1 

coty  +  cot  a; 
cot  a;  cot?/  +  1 

coty  —  cot.c 

(I) 


X 

90°— X 

90° +x   ;  180°— a; 

1 

180°  +  x 

270°—  X 

270°  +  X 

360°- a; 

360° +x 

tfhie 

cos  a; 

cos  a; 

sin  a; 

—  sin  X 

—  cos  a; 

—  cos  a; 

—  sin  X 

sin  a; 

uoeine 

sin  a; 

—  sin  a; 

—  cos  a; 

—  cos  X 

—  sin  X 

sin  X 

CO:?  a; 

cos  a; 

tangent 

cot  a; 

—  cot  a; 

—  tana; 

tanx 

cot  a; 

—  cot  a; 

—  tana; 

tana; 

cotangent 

tan  X 

—  tana; 

—  cdt  x 

cot  X 

tan  X 

—  tan  X 

—  cot  a; 

cot  a; 

It  will  not  be  found  difficult  to  memorize  and  extend  set  (I),  if  the  student 
observes,  that,  when  the  number  of  whole  quadrants  is  odd  (as  90°,  270°,  etc.), 
the  function  changes  name  (as  from  sin  to  cos,  from  cos  to  sin,  etc.) ;  but,  when 
the  number  of  whole  quadrants  is  even,  the  function  retains  the  same  name. 
The  sign  of  the  sine  and  cosine  is  readily  deteiTiiined  according  to  fundamental 
principles  by  observing  where  the  arc  ends,  assuming  x  <  90°.  Thus  180°+  x 
ends  in  the  third  quadrant ;  hence  its  sine  (which  in  numerical  value  is  sin  a*)  is 
—  ,  and  its  cosine  is  also  — .  As  the  signs  of  the  tangent  and  cotangent  of  the 
sOi^ne  arc  are  alike,  we  have  only  to  observe  whether  the  sine  and  cosine,  in  any 
tjiven  case,  have  like  or  unlike  signs,  in  order  to  determine  the  sign  of  the  tan- 
gent and  cotangent.  For  example,  what  is  cot  (630°  +  a; )  equal  to  ?  The  num- 
ber of  quadrants  being  odd  (7),  the  function  changes  naj)ie,  and  since  the  arc  ends 
in  the  fourth  quadrant,  its  sine  is  — ,  and  its  cosine  +  ;  therefore  cot  (630°  +  x) 
=  —  tan  x.  If  in  any  given  case  x  >  90°,  determine  the  character  of  the  function 
as  above,  on  the  hypothesis  x  <  90°,  and  then  modify  the  result  for  the  partic- 
ular value  of  X.  Thus  in  the  last  case,  if  x  was  between  90°  and  180",  its  tan- 
gent would  be  — ,  and  for  such  a  value  cot  (630°  +  x)  =  tan  x.  Or,  we  may 
consider  at  first  where  the  arc  ends,  taking  into  consideration  the  given  value 
of  a;. 


26 


PLANE   TRIGONOMETRY. 


(b)    FUXCTIOXS   OF   DOUBLE   AND   HALF   ANGLES. 

30,  Pr02>'  — Letting  x  represent  any  angle  {or  arc), 


(K)  siu  2x  =  2sin  x  cos  x ; 

(L)  cos  2x  =  cos' a;  —  sin'' a:  = 
^^      2cos"  X  —  1,  or  1  —  2siu''  x ; 


(M)  tim2x 
(N)  cot2.'c 


2  tan  x 
1  -  tan'  a 
cot'  X  —  1 

2cot  X 


Dem.  These  results  are  readily  deduced  from  (A),  (C),  (E),  aud  (G)  Thus,  in 
siu  {x  +  y)  =  sin  x  cos  y  +  cos  x  sin  y,  if  we  make  y  —x^  we  have  sin 2x  = 
Bin  X  cos  a;  +  cos  x  siu  x  =  2siu  x  cos  a;.    (In  like  manner  produce  the  others.) 


o7.  J^rox), — Letting  x  rejjresent  any  angle  (or  arc),  we  have, 


(0)  sinfr  =  ±Vi(l-cosa;); 


(P)  cos  \x  =  ±  Vi  (i  +  cos  a;) ; 


(Q)  tan  i.  =  ± /l^^^^ 
1  +  COS  a; 


(R)  cot  J  a; 


•^^ 


cos  a; 


cos  a; 


Dem.    From  (L),  2sin"''a;  =  1  -  cos  2^-,  or  sin  a;  —  ±  /y/i  (1  —  cos  2a;).    Puttiug 

ix  for  X,  this  becomes  siu  ia;  =  ±  -y/i  (1  —  cos  x).   In  like  manner,  from  the  same 

formula  (L),  2cos^a;  =  1  +  cos  2x ;   whence,  cos  \x  =  ±  ^\  (1  +  cos  x\   Again, 

.  „.         sini.r  /l  — cosa;        .        .  1  ,  /l  +  cos  x 

tan \x  = =  ±i/ ;  and  cot^a;  = —  =  ± i/ .    Q.  E.  D. 

cosic  '    1  +  cosa;  tan  ^a;  r    i  —  cosa; 

ScH.     The  sign  of  the  function  in  the  case  of  each  of  these  is  +  if  a;  <  180"  ; 
but  can  only  be  determined  b}'  the  value  of  a;  in  any  given  case. 


EXERCISES. 

1.  Prove  from  Fig.  (a)  that  sin  {x  -f-  y)  =  sin  x  cos  y  +  cos  x  sin  y, 
when  X  and  y  are  each  <  90°,  but  x  +  y 
>  00°. 


2.  Same  as  in  £Jx.  1,  from  (b),  when 
X  <  90°,  X  -\-  y>  90°,  and  <  180°,  and 
y  >  90°  and  <  180°. 

SuG.  hi  this  case,  sin  (x  -^  y)  =  P'D'  =  P'L  - 
EF.  Ill  other  respects  the  demonstration  is 
identical  with  the  preceding.  This  gives 
sin(a;  +  y)  =  cos  a;  sm  y  —  sin  a;  cos  y.    But  the 


0  F     D      Ci^A. 


(a). 


^u^"CTIONS  of  the  sum  or  difference  of  angles. 


27 


0). 


—  sign  is  accounted  for  in  the  general  formula, 
sin  {x  +  y)  =  sin  a;  cos  y  +  cos  x  sin  y,  by  notic- 
ing that  cosy  is  — ,  when  y  >  90°  and  <  180°, 

3.  Same  as  in  the  preceding,  when 
X  >  90°,  y  <  90°,  and  (x  +  y)  <  180°. 

Sua.  Here  sin  (.2;  +  2/)  =  P'D'  =  EF  -  P'L  In 
all  other  respects  the  demonstration  is  identical 
with  the  otlier  cases.  The  —  sign  in  this  case 
arises  from  x  being  between  90'  and  180", 
whence  cos  a;  is  — . 

[Note.  A  number  of  other  cases  may  be 
devised,  but  the  ri)ove  illustrate  the  varieties.] 

4.  From  i'V^.  11,Art.  48,  demonstrate 
geometrically  the  formula  cos  {x  ■{■  y)  = 
cos  X  cos  ?/  —  sin  a;  sin  y.  The  same  for 
each  of  the  cases  in  Ex's  1,  2,  and  3, 
above. 


SuG.     In  Fig.  11,  cos  {x  +  y)  =  OD'  =  OF  - 
LE.     g|  =  ^^,  or  OP  =  cos  x  cos  2/-  -^  =  q^,  or  LE  =  sin  x  sin  y, 

5.  Prove  geometrically  the  relation  sin  [x  —y) 
=  sin  X  cos  y  —  cos  x  sin  y. 

SuG.  Let  aP  =x,  and  since  y  is  to  be  subtracted  we 
measure  it  back  from  P,  and  y  =  PP'.  Now  sin  {x  —  2^) 
=  P'D'=r  EF  -  P'L. 

0.  Prove  from  Fig.  (d)  that  cos  {x  —  y)  = 
cos  X  cos  y  +  sinx  sin  y. 


DaA 


7.  Given  sin  45°=  Vi^  and  sin  30°=  J  to  find  sin  75°,  and  sin  15" 
Also  tan,  and  cot.     Result^   sin  75°  =  .97,  sin  15°=  .2G,  nearly. 

SuG.    Use  formula  (A  ,  .  .  .  H). 

8.  Given  sin  30°=  J,  to  find  sine,  cosine,  tangent,  and  cotangent, 
of  15°,  7°  30',  3°  45',  and  1°  52' 30".  Results,  sin  15°=  .2588,  cos  15° 
=  .97,  nearly. 

SuG.  Use  the  formulm  in  (57).  Compare  results  with  those  fcnmd  in  tha 
Table  of  Natural  Sines,  etc. 

9.  Of  what  angles  may  the  trigonometrical  functions  be  found 
from  sin  45°=  jV2,  by  means  of  the  fornuihe  in  (J7)  ?    How  ? 


28  PLANE   TRIGONOMETRY. 

10.  Prove  that  sin  (x  -hy  +  z)  =  sin  x  cos  y  cos  z  +  cos  x  sin  y  cos  z 

+  cos  a;  cos  y  sin  2  —  sin  x  sin  y  sin  2;.      Also.,  cos  {x  -\-  y  +  z)  = 

cos  a;  cos  y  cos  2;  —  sin  x  sin  ycosz  —  sin  a;  cos  ?/  sin  2;  —  cos  x  sin  ?/  sin  z. 

. ,      , ,    ,  ^      /  .       tan  X  +  tan  y  +  tan  z  —  tan  a:  tan  y  tan  2 

Also  that  tan  (x-{-  y  +z)=: —^ —^ . 

^  1  —  tan  X  tan  y  —  tan  x  tan  z  —  zany  tan  z 

SuG.    Sin  {x  +  y  +  z)  =  sin  [{x  +  y)  +  z]  =  sin  (.2;  +  2/)  cos z  +  cos  (a;  +  y) sin z. 

ScH.  Since  if  {x  +  y  +  z)=  Tt^  tan  (^  +  y  4-  2)  =  0,  we  have  from  the  last 
form,  tana;  +  tan y  +  tans  =  tan x  tan  y  tan z;  i.  e.,  if  a  semicircumference  be 
divided  into  any  three  parts,  the  sum  of  the  tangents  of  the  three  parts  equals 
the  products  of  the  tangents  of  the  same. 

^  ^    T^  . ,    ,         ,  V         sec  a:  sec  y  cosec  x  cosec  y 

11.  Prove  that  sec  (x  +  y)  = —, 

^ '       cosec  X  cosec  y  —  sec  x  sec  y 

12.  Prove   that    sin  dx  =  3sin  x  —  4sin^  x.      Also  that  cos  3x  = 

4:Cos^  X  —  3cos  X.    Also,  tan  3.?;  =  — :; ^z — '-^ — -»     Also,   cot  3a;  = 

1  —  3  tan  a; 

cot' a;  —  3cota; 


3cot'  a;  -  1 

SuGS.  Sin  3^  =  sin  {2x  +  x)  =  sin  2x  cos  x  +  cos  2x  sin  x  =  2sin  x  cos  x  cos  a; 
+  (1  —  2sin'^)  sin  x  =  2sin  x  cos'  a;  +  sin  a;  —  2sin'  x  =  2sin  a;  (1  —  sin''  x)  +  sin  aj 
—  2sin'  X  =  Ssin  x  —  4sin''  x. 

tan  '^  c  +  tan  a^ 

Tan  dx  =  tan  (2.c  +  a;)  =  :; ^^^— .   In  the  latter  substitute  for  tan  2a;  its 

1  —  tan  2a;  tan  x 

value  iu  terms  of  tan  x. 

13.  Prove  that  sin  4a:  =  4(sin  x  —  2sin''a;)  cos  a;. 

lA    r>  i-i    r    •  2tan4a;  2 

14.  Prove  that  sm  x  = T—^-r-  = z — r-. 

1  +  tan  ^x      cot  ^x  +  tan  ^x 

sin  hx 


„         „.  r.  •     ,  ,         2smir         cos^a;      2tan  i.r        2tan  i.?;       ^ 

SuG.    Sm  X  =  2sm  ^c  cos  ^v  = f-  = p-  =  — ^-^  =  , — — -4-r .    To 

sec  i.c        sec  i-v       sec'  i  a;      1  +  tan'  i^x 


produce   the  last  form   divide  numerator    and  denominator    of  t-^^^ It- by 

1  +  tanHa;   "^ 


1  -    T>  i-i    i.  i.       1         1  —  cos  a;       .  -  ,  ^         1  4-  cos  a; 

lo.  Prove  that  tan  la;  =  — ■. .     Also  cot  la;  = : . 

sin  a;  "  sm  x 

SuG.   From  (L),  (56*),  2sin''  ^x  =  l  —  cos  x,  and  from  (K),  2sin  \x  cos  \x  =  sin  x. 
Divide  the  fornior  by  the  latter. 


FORMULiE  ADAPTED   TO   LOGAEITHAIIC   COMPUTATION.  2^ 

16.  Find  the  trigonometrical  functions  of  18°. 

Solution.— Letting  x  =  18°,  2x  =  3G°,  and  3x  =  54°,  hence  sin  2x  =■-  cos  Sx 
But  sin  2x  =  2sin  x  cos  x ;  and  cos  Zx  =  4005"  x  —  3cos  x ;  hence  2sin  a;  cos  ^  = 
4cos'  X  —  3cos  a;,  or  2sin  x  =  4cos'  x  —  3  =  4t  —  4siu'  a;  —  3.    From  which  4sin'  x 

+  2sin  x  =  l.     Solving  this  quadratic,   we  have  sin  x,  or  sin  18°  =  -^^— j — 


neglecting  the  —  root,  since  sin  18°  is  +.    From  this,  cos  18°  =  \\/^^  +  2/y/5. 
These  may  be  put  in  approximate  decimal  fractions. 

17.  Having  given  the  functions  of  18°,  and  15°  (Ex.  8),  find  those 
of  3°  ;  then  of  6°,  12°,  24°,  etc. 

Compare  the  results  obtained  with  the  values  as  given  in  the  Table  of  Natural 
Sines,  etc.,  obtaininG;  all  the  values  in  decimal  fractions. 


SUCTION  III 

FORMULIE  FOR  RENDERING  CALCULABLE  BY  LOGARITHMS  THE 
ALGEBRAIC  SUM  OF  TRIGONOMETRIC.iL  FUNCTIONS. 

S8,  Since  multiplication,  division,  involution,  and  evolution  are 
the  only  elementary  combinations  of  number  which  we  can  effect  by 
means  of  logarithms,  if  we  l^isli  to  add  or  subtract  trigonometrical 
(or  other)  quantities,  we  have  first  to  discover  what  products, 
quotients,  powers,  or  roots,  are  equivalent  to  the  proposed  sums  or 
difierences. 


SO*  IPvop, — To  render  sin  x  ±  sin?/,  and  cos  a;  db  cos  y  calculable 
by  logarithms. 

Solution.  From  {55,  Son.  2)  we  have  sin  (2;  +  y)  =  sina;  cosy  +  cos  a: 
siny,  and  sin  {x  —  y)  —  sin  x  cos  y  —  cos  x  sin  y.  Adding  these  formulas 
sin  {x  +  y)  +  sin  {x  —  y)  =  2sin  x  cos  y.  Now  putting  x  +  y  =  x\  and  x  —  y  =  y'  • 
whence  x  =  ^{x'  +  y'),  and  2/  =  i  {x'  —y') ;  we  have  sin  x'  +  sin  y'  =  2sin  l{x'  +  y') 
cos^(a;'  —  y);-or,  dropping  the  accents,  as  the  results  are  general, 

(A')  sin  X  +  s'my  =  2sin  ^{x  +  y)  cos  \{x  —  y). 

Again,  by  subtracting  formula  B  {55,  Son.  2)  from  formula  A,  and  makuig 
'he  same  substitutions,  we  have, 

(B')  sin  X  —  siny  =  2cos  ^{x  +  y)  sin  i{x  —  y). 


30  PLA^'E  TRIGONOSrETET. 

In  like  manner  adding  cos  {x  +  y)  =  cos  a:  cos  y  —  sin  x  sin  y,  and  cos  {x  —  y)  = 
cosx  cosy  +  sinar  siuy,  and  making  the  same  substitutions,  we  have, 

(C)  cos X  +  cosy  =  2cos \{x  +  y)  cos  \{x  —  y). 

Finally  subtracting  formula  C  (55,  ScH.  2)  from  formula  D,  and  making 
the  same  substitutions,  we  have, 

(DO  cos  y  —  cosx  =  '    2sin  ^{x  +  y)  sm  \{x  —  y\  or 

cos x  —  co5y=  —  2sin l{x  +  y)  sin i{x  —  y). 

00.  Cor.  l. — T7te  su?yi  of  the  sines  of  two  angles  is  to  tlieir  differ- 
ence, as  the  tangent  of  one-half  the  sum  of  the  angles  is  to  the  tangent 
of  one-half  their  diff'erence. 

Dem. — Dividing  A'  by  B',  we  have, 

sin  X  +  sin  y  _  sin  \{x  +  y)  cos  \{x  —  y)  _  sin  \{x  +  y)     cos  i{x  —  y)  _ 

sin «  — sin  y  ~  cos  \{x  +  y)  sin  \{x  —  y)~  cos  \{x  +  y)      sin  \{x  —  y)~ 

1  tin  ^(x  -\-  v) 

tanKx  +  y)coti(.r-y)  =  tani(^  +  y)  x  __^_-_^  = -1-|  ___   q.  e.  d. 

61»  Cor.  2. — TJie  difference  of  the  cosines  of  two  angles  divided  by 
tlieir  sum  is  numerically  equal  to  the  product  of  the  tangent  of  one- 
lialf  the  Slim  of  the  t2co  angles  into  the  tangent  of  one-half  their 
difference. 

Dem. — Dividing  D'  by  C,  we  have, 

cos  a;  —  cosy  _  —  sin  \{x  +  y)  sjn  ^x  —  y)  _  _  sin  \{x  +  y)      sin-^(3;  —  y) 

cos  a:  +  cos  y  ~  cos  \{x  +  y)  cos  l{x  —  y)    ~      cos  \(x  +  y)      cos  \{x  —  y)  ~  • 
tan  \{x  +  y)  tan  \{x  —  y).    q.  e.  d.    (Observe  the  opi)osition  in  signs.) 


02,  JProb, — To  render  tan  x  db  tan  f/  calciilahle  by  logarithms. 

_  _  ^  sin.?;       sin  V      sin  t  cos  y  ±  cos  a?  sin  y      sin(a:±y) 

Dem.— Tan  a;  ±  tan  y  = ±  ^  = — — 

cos  a:      cosy  cos  a,- cosy  cos  a;  cosy 

q.  E.  D. 


EXERCISES. 

Let  the  student  deduce  the  following  relations : 

-.    i-.  i.      .       ^  sin  {x  +  ?/) 

1.  Cot  X  +  coty  =  -T— 5^ — 7-^. 

^        sin  X  sm  y 

«    o  2  cos  i(.T  +  y)  cos  i(:c  —  y) 

2.  Sec  a;  +  sec  y  = ^ ^ ^ -. 

^  cos  X  COS  y 

«    o  2  sin4-(a:  +  ?/)  sini(.'C  —  y) 

3.  Sec  a;  —  secy  = — — -- 

^  COS  a;  cosy 


COXfeTRUCTION   AND   USE   OF   TIlIGONOMETRlCAL   TABLES.  31 

4.  1  +  COS  a;        =  2cos'ia;.     (See  S7.) 

5.  1  —  cos  a;        =  2sin'Jrc. 

6.  £!L^-iHll^  =  tani(a;  +  y).  .  (Divide  A'byC^^.9.) 
cos  re  +  cos?/  ^^  ^/       V  J      7         / 

^    sin  a:  —  sin  y       ,       , ,          . 

7.  ; =  tan  i(3;-  t/). 

cos  a:  +  cosy  ^v       .// 

_    sin  ic  +  sin  ?/  ,  ^  .  . 

8. =  —  cot  i(a;  —  y). 

cos  a;  — cos?/  *^        ^^ 

^    sin  a:  —  sin  V  '     ,  , , 

9.  ^  =  —  cot  -Ka;  +  y). 

cos  a;—  oosy  '^^        '" 


SECTION  IV. 

CONSTRUCTION  AND  USE  OF  TRIGONOMETRICAL  TABLES. 

[Note. — In  order  to  read  this  and  the  subsequent  sections,  the  student  needs 
a  knowledge  of  the  nature  of  logarithms,  and  the  method  of  using  common 
logarithmic  tables.  If  he  is  familiar  with  the  last  chapter  in  The  Complete 
School  Algebra  of  this  series,  he  is  j^repared  to  go  on.  If  lie  has  not  this 
knowledge,  he  should  read  the  inti'oduction  preceding  the  table  of  Logarithms 
before  reading  this  section.] 

63,  A  Table  of  Trigonometrical  Functions  is  a  table 
containing  the  valnes  of  these  functions  corresponding  to  angles  of 
all  different  values.  In  consequence  of  the  incommensurability  of 
an  arc  aAd  its  functions,  these  results  can  be  given  only  approxi- 
mately; yet  it  is  possible  to  attain  any  degree  of  accuracy  which 
practical  science  requires. 

04:,  There  are  two  tables  of  trigonometrical  functions  in  common 
use,  the  Table  of  Natural  Functions,  and  the  Table  of  Logarithmic 
Functions. 

05,  A  Table  of  JS'atiiral  Trigonometrical  Functions 
is  a  table  in  which  are  written  the  values  of  these  functions  for 
angles  of  various  values,  the  radius  of  the  circle  being  taken  as  the 
measuring  unit,  and  the  function  being  expressed  in  natural  num- 
bers extended  to  as  many  decimal  places  as  the  proposed  degree  of 
accuracy  requires. 

00,  A  Table  of  Logarith^nic  Trigonometrical  Func- 
tions is  the  same  as  a  table  of  natural  functions,  except  that  the 
logarithms  of  the  values  of  the  functions  are  written  instead  of  the 
functions  themselves,  and  to  avoid  the  frequent  occurrence  of  nesja- 


32  %         PLANE  TTJGONOMilTEY. 

tive  characteristics,  the  characteristic  of  each  logarithm  is  increased 
by  10.  For  example,  sines  and  cosines  being  always  less  than  unity, 
except  at  the  limit  (33),  and  tangents  of  angles  less  than  45°  and 
cotangents  of  angles  greater  than  45°  being  also  less  than  unity,  the 
logarithms  of  all  such  functions  have  negative  characteristics.  To 
obviate  the  necessity  of  writing  these  with  their  sign,  the  charac- 
teristic of  each  logarithm  is  increased  by  10. 


67,  JProb, — To  compute  a  table  of  natural  trigonometrical  func- 
iionsfor  every  degree  and  minute  of  the  quadrant. 

Solution. — It  is  evident  that  an  arc  is  longer  than  its  sine,  but  that  this 
disparity  diminishes  as  the  arc  grows  less.  Thus,  in  a  circle  whose  radius  is 
1  inch,  the  length  of  the  sine  of  an  arc  of  1°  would  not  difer  appreciably  from 
the  arc.  Much  less  should  we  be  able  to  distinguish  between  the  sine  of  1'  and 
the  arc.  Now,  since  when  the  radius  is  1,  a  semicircumference  =  7t  =  3.141592G, 
and  also  =  180',  or  ISO  X  GO  =  10800',  we  have  the  length  of  an  arc  of  V  = 

"^  =  0.0002908882  approximately.     Assuming  this  as  the  sine  of  1',  we 
lObOO 

obtain  the  cosuie  thus. 


cos  1'  =\/l  -  sin'^  r  =\/(l  +  sinl')  X  (1-sinl')  =  v/l.0002908882  x  .9997091118 
=  0.9999999577. 

Having  thus  obtained  sufficiently  accurate  values  of  sin  1'  and  cos  1',  we  can 
continue  the  operation  as  follows:  from  the  formula  sin  (a:  +  y)  +  sin {x  —  y)  = 
2  sin  X  cos  y,  and  cos  {x  +  y)  +  cos  {x  —  y)  =  2  cos  x  cos  y,  we  have 

sin  (x  +  y)  =  2  sin  ar  cos  y  —  sin  {x  —  y),  ' 

cos  (x  +  y)  =  2  cos  a;  cos  y  —  cos  [x  —  y). 

Now  letting  y  remain  constantly  equal  to  1',  and  letting  x  take  successively 
the  values  1',  2',  3',  etc.,  we  have 

_        j  sin  2'  =  2  cos  1'  sin  1'  -  sin  0'  =  0.0005817764 
For  a:  _  1  ,   ^  ^^^  ^^  _  ^  ^^^  ^,  ^^^  ^  _  ^^^  q,  _  0.9999998308 

_  ^,    j  sin  3'  =  2  cos  1'  sin  2'  -  sin  1'  =  0.000872664G 
For  ar  -  ~  ,   ^  ^^^^  3'  =  2  cos  1'  cos  2'  -  cos  1'  =  0.999999G193 

_    ,  I  sin  4'  =  2  cos  1'  sin  3'  —  sm  2'  =  0.001 1G35526 

For  ar  _  3  ,  ^  ^os  4'  =  2  cos  1'  cos  3'  -  cos  2'  =  0.9999993232 

_    ,  (  sin  5'  =  2  cos  1'  sin  4'  -  sin  3'  =  0.0014544407 

For  X  -  4',  -^  cos  5'  =  2  cos  1'  cos  4'  -  cos  3'  =  0.9999989425 
etc.,  etc. 

These  operations  present  no  difficulties  except  the  labor  of  performing  the 
numerical  operations. 

Of  course  GO  operations  are  required  for  ever}'  degree,  and  for  30°,  1800.  But 
having  computed  the  sines  and  cosines  for  every  degree  and  minute  up  to  30° 


CONSTRUCTION  AND  USE  OF  TRIGONOMETRICAL  TABLES.    33 

we  can  complete  the  work  by  simple  subtraction  of  values  already  fouml     For 
example,  letting  x  =  30°,  the  first  formula  used  above  becomes 

sin  (30°  +  y)  =  cos  y  -  sin  (30°  -  y\ 
and  from  cos  {x  +  y)  —  cos  {x  —  y)=  —  2siD  x  sin  y,  we  have 

cos  (30' +  y)  =  cos  (30°  —  y)  —  sin  y. 
Now  making  y  successively  =  1',  2',  3',  etc.,  these  give 

j  sin  30*  V  =  cos  1'  —  sin  29°  50 

(  cos  30°  1'  =  cos  29°  59'  -  sin  1' 

j  sin  30°  2'  =  cos  2'  -  sin  29°  5S' 

i  cos  30°  2'  =  cos  29°  58'  -  sin  2' 

j  sin  30°  3'  =  cos  3'  -  sin  29°  57' 

(  cos  30°  3'  =  cos  29°  57'  —  sin  3' 
etc.,  etc. 

All  of  these  values  which  occur  in  the  second  members  having  been  deter- 
mined in  reaching  sin  30'  and  cos  30°,  those  in  the  first  members  can  be  found 
by  performing  the  requisite  subtractions. 

Proceeding  in  this  way  till  we  reach  45°,  the  numerical  values  o^  all  sines  and 
cosines  become  known,  since  the  sine  of  any  angle  between  45°  and  90°,  being 
the  cosine  of  the  complementary  angle,  will  have  been  computed  in  reaching 
45°.  And  so  also  the  cosines  of  angles  between  45°  and  90°  will  have  been 
computed  as  sines  of  the  complementary  angles  l^elow  45°. 

The  sines  and  cosines  being  computed,  the  corresponding  tangents,  cotan- 
gents, and,   if  need   be,  the   secants,   cosecants,  versed-sines,   and   coversed- 

sin  2  1.  cos  7! 

sines,  can  be  calculated  from  the  relations  tan  x  = ,  cot  x  = or  -: — ~, 

cos  X  tan  c       sm  x 

1  1  .  ,  .        . 

seca;  = ,  cosecx  =  - — ,  vers  x  =  1  —  cosa*,  and  covers  .c  =  1  —  sm  .r. 

cos  X  sm  j; 

68,  Sen. — If  it  is  desired  to  obtain  the  natural  functions  of  angles  esti- 
mated to  seconds,  it  is  necessary  that  the  values  in  the  ta])les  computed  as  above 
be  extended  to  7  decimals  at  least.  From  such  a  table  we  may  make  iiiteri)o- 
lations  for  seconds  with  suflScient  accuracy  for  most  practical  ends,  except 
for  values  near  the  limits,  where  the  disparity  between  the  variation  of  the  aic 
and  that  of  the  function  changes  very  rapidly.  For  example,  let  it  be  required 
to  find  sin  34°  24'  12  "  from  the  data  sin  34°  24'  =  .5G49G70,  and  sin  34°  25'  = 
.5652070.  We  observe  that  an  increase  of  1'  upon  the  angle  of  34°  24'  makes 
an  increase  of  .5652070  —  .5649670  =  .0002400  in  the  sine.  Hence  an  increase 
of  12",  or  i  of  1',  makes  an  increase  of  i  of  .0002400,  or  .0000480,  approxi- 
mately. Adding,  we  have  sin  34°  24'  12"  =  .5650150.  The  student  must  be  careful 
to  notice  whether  an  increase  of  the  angle  makes  a  numerical  increase  or  a  de- 
crease of  the  function,  and  add  or  subtract  as  the  case  may  require. 


69.  JProb, — To  constrtict  a  table  of  logarWimic  trigonometrical 

functions. 

Solution. — Compute  the  natural  sines  and  cosines  as  in  the  preceding  prob- 
'em.    Take  the  h^garithms  of  the  values  tlius  obtained,  and  add  10  to  each 

3 


34:  PLANE  TRIGONOMETRY. 

cLaractcristic.  The  results  are  the  ordinary  tabular  logarithmic  sines  and  co- 
sines. For  example,  we  find  from  the  table  of^natural  functions  that  sin  34°  24' 
r-  .5049670.  The  logarithm  of  this  number  is  1.752023.  Adding  10  to  the  char- 
acteristic, Tre  have  log  sin  34'  25'  =  9.752023,  as  usually'-  given  in  the  tables.  In 
like  manner  the  cosines  are  obtained. 

To  obtain  the  tabular  logarithmic  tangents,  we  have  from  tan  x  — 

cos  X 

log  tan  X  =  log  sin  x  —  log  cos  x.     If  we  now  take  the  log  sin  x  from  the  table  as 

computed  by  the  preceding  part  of  this  solution,  and  from  it  subtract  the  cor- 

/espondlng  log  cos  ar,  the  result  is  tlie  true  log  tan  x^  since  the  extra  10  in  the 

tabular  log  sin  and  log  cos  is  destroj^cd  by  the  subtraction.    Therefore,  to  thi« 

difference  we  must  add  10  to  get  the  tabular  log  tan,  as  above  explained.    For 

example,  the  tabular  log  sin  34°  24'  =  9  752023,  and  log  cos  34°  24'  =  9.916514. 

Hence,  the  tiibular  log  tan  34°  24'  =  9.752023  -  9.916514  +  10  =  9.835509.    In 

like  manner  the  tabular  log  cot  x  =  log  cos  x  —  log  sin  x  +  10. 

If  the  logarithmic  secants  are  required  they  can  be  obtained  from  the  relation 

sec  x  = ,  which  gives  log  sec  :c  =  0  —  log  cos  x.    In  applying  this  by  means 

cos.^ 

of  tlie  tabular  functions,  it  must  be  observed  that  the  log  cos  x,  as  we  get  it 
from  the  table,  is  10  too  great ;  hence,- the  tnte  log  sec  a;  =  0  —  log  cos  a;  +  10. 
In  tabulating  log  secants  and  cosecants,  it  is  not  necessary  to  add  10,  since,  as 
these  functions  are  never  less  than  1,  their  logarithms  are  never  negative. 

70.  Sen. — The  interpolations  for  seconds  are  usually  made  in  the  same  way 
when  using  the  logarithmic  functions,  as  explained  above  for  the  natural  fimc- 
tions.  But  to  facilitate  the  operation,  the  approximate  change  of  the  logarithm 
for  a  change  of  1"  of  the  angle  is  commonly  written  in  the  table,  in  a  colmnn 
called  Tabular  Differences,  and  marked  D. 


EXERCISES. 

1.  Find  from  the  tables  cat  the  close  of  the  vohime  the  natural 
trigonometrical  functions  of  25°  18'. 

Solution. — To  find  (he  sine  and  cosine  w^  look  in  Table  II.,  and  find  25°  at 
the  top  of  the  page.  In  the  extreme  left-hand  column  we  find  the  minutes,  and 
passing  down  to  18,  find  opposite,  in  the  column  headed  N.  sin  (natural  sine) 
42736 ;  also  in  the  column  N.  cos,  we  find  90408.  Now,  as  these  are  the  lengths 
of  the  sine  and  cosine  as  compared  with  radius,  we  know  they  are  fractions. 
.-.  Sin  25°  18'  =  .42736,  and  cos  25°  18'  =  .90408. 

To  find  the  tangenty^e  turn  to  Table  IV.,  and  finding  25°  at  the  top  of  the  page, 
pass  down  the  column  of  minutes,  on  the  left-hand  of  the  page,  to  18,  opposite 
which,  and  under  the  column  headed  25°,  we  find  2698.  To  this  we  prefix  the 
figures  47,  which  stand  in  the  same  column,  opposite  11',  and  belong  to  the  tan- 
gents of  all  tlie  angles  from  25°  10'  to  25°  19',  and  are  omitted  in  the  table  sim- 
ply to  relieve  the  eye  and  to  economize  space.  Thus  we  find  tan  25°  18'  =  .472698, 
tha  number  being  laiown  to  be  a  fraction  because  the  angle  is  less  tlian  45° 


CONSTRUCTION  AND  USE  OF  TRIGONOMETRICAL  TABLES.    oO 

Tojincl  the  cotangent  we  look  at  the  bottom  of  the  page  in  the  same  table  till 
we  find  25*,  and  tlien  passing  up  the  minutes  column  at  the  right  hand,  find 
c:ot25°  18' =  2.11552. 

If  the  aecant  were  required  we  should  be  obliged  to  obtain  it  by  dividing  1  by 
the  cosine,  as  our  tables  do  not  include  this  function.    Thus  sec  25°  18'  = 

cos  25°  18'  ^  ~Mm  ^  ^••^^^^• 

[Note. — Tables  of  secants  and  cosecants  are  sometimes  given,  but  they  are 
not  of  sufficient  importance  to  justify  their  introduction  into  an  elementaiy 
text-book.] 

2.  Show  that  sin  37°  43'  =  .61176 ;  cos  37°  43'  =  .79105 ;  tan  37°  43' 
=  .773353;  cot  37°  43' =  1.29307;  sec  37°  43'=  1.2G4142;  cosec37° 
43'  =  1.634628 ;  yers  37°  43'  =  .20895 ;  covers 37°  43'  =  .38824. 

3.  Find  that  sin  64°  36' =  .90334;  cos  64°  36' =  .42894;  tan  64° 
36'  =  2.10600 ;  cot  64°  36'  =  .474835 ;  sec  64°  36'  =  2.331328 ;  coseo 
64°  36'  =  1.107003 ;  vers  64°  36'  =  .57106;  covers  64°  36'  =  .09666. 

SuG. — In  looking  for  sines  and  cosines  of  angles  above  45°,  seek  the  degrees 
at  the  bottom  of  the  page,  and  be  careful  to  observe  that  the  columns  of  sines  and 
cosines,  as  named  at  the  top,  change  names  when  read  from  the  bottom.  The 
foundation  of  this  arrangement  will  be  readily  perceived.  Thus,  turning  in 
Table  II.  to  24°  32',  we  find  sin  24°  32':^  .41522.  But  sin  24°  32'  =  cos  (90  -  24°V2') 
=  cos  65°  28'  =  .41522.  Thus  the  degrees  and  minutes  read  from  the  bottom  of 
the  page  are  the  complements  of  those  read  from  the  top. 

4.  Find  that  sin  42°  27'  12"  =  .67499;  cos  42°  27' 12"  =  .73783; 
tan  42°  27'  12"  =  .914834;  cot  42°  27'  12"  =  1.09309. 

SuG.— Sin  42°  27'  =  .67495,  and  sin  42°  28'  =  .67516.  .'.  An  increase  of  1'  in 
the  angle  makes  an  increase  of  21  (hundred-thousandths)  in  the  sine,  and  12" 
will  make  ^^  or  \  as  great  an  increase,  approximately.  Observe  that  in  the  case 
of  cosine  an  increase  of  the  arc  makes  a  decrease  of  the  function. 

5.  Find  that  sin  143°  24'  =  0.596225;  cos  151°  23' =  .877844; 
tan  132°  36'  =  1.08749;  and  cot  116°  7'  =  .490256. 

SuG.— Sin  143"  24'  =  sin  (180°  —143°  24')  =  sin  30°  36'.  Also  the  trigone- 
metrical  function  of  any  angle  is  numerically  equal  to  the  same  function  of  its 
iiupplement  {56). 


6.  Find  the  logarithmic  trigonometrical  functions  of  32°  15'  2; 
from  the  tables  at  the  end  of  the  volume. 


36  PLANE   TRIGONOMETRY. 

SOLUTION.— Turning  to  Table  II.  we  find  32°  at  the  top  of  the  page,  and 
opposite  15',  and  in  the  column  L.  sin  (logarithmic  sine),  we  get  9.727228  ;  i.  e., 
log  sin  32°  15'  =  9.727228.  Kow  from  the  column  of  differences,  D.  1",  we  learu 
that  an  increase  of  1"  of  the  arc  at  this  point  makes,  approximately,  an  increase 
of  3.34(raillionths)  in  the  logarithm  of  its  sine.  Hence,  we  assume  that  an  in- 
crease of  22"  makes  22  x  3.34  =  73  (millionths).  .-.  log  sin  32°  15'  22"  =  9.727228 
-f  .000073  =  9.727301.  In  a  sunilar  manner  we  have  log  cos  32°  15'  =  9.927231. 
An  increase  of  1"  in  the  arc  makes  a  decrease  of  1.33  (millionths)  in  the  log  cos. 
.-.  an  increase  of  22"  makes  29  (millionths)  decrease  in  the  log  cos,  and  log  co? 
82=  15'  22"  =  9.927202.  Log  tan  32°  15'  22"  =  9.800100 ;  and  log  cot  32'  15'  22"  = 
10.199901. 

7.  Find  rliat  log  sin  24°  27' 34"  =  9.617051 ;  log  cos  26°  12'  20"  = 
9.952897 ;  log  tan  26°  12'  20"  =  9.692125  ;  log  cot  126°  23'  50"  = 
9-8675;9. 

Sufe.— Observe  cot  (126^  23'  50")  =  cot  (180°  -  126°  23'  50")  =  cot  (53°  36'  10"). 
Also  that  angles  above  45'  are  found  at  the  bottom  of  the  table ;  and  remember 
to  subtract  the  correction  for  co-functions,  if  an  increase  of  arc  is  assumed. 


8.  Given  the  natural  sine  .45621,  to  find  the  angle  from  the  tables. 

Solution. — Looking  for  this  sine  in  the  table  of  natural  sines,  we  find  the 
next  less  sine  to  be  .45606,  and  the  angle  corresponding,  27°  8'.  Now,  at  this 
point,  an  increase  of  V  in  the  arc  makes  an  increase  of  20  (hundred  thousandths) 
in  the  natural  sine.  But  the  given  sine  .45621  is  only  15  (hundred  thou- 
sandths) greater  than  .45606,  the  sine  of  27"  8'.  Hence  the  required  angle  is 
but  H  of  V  or  60"  =  35",  greater  than  27'  8'.  .-.  sin-».45621  =  27°  8'  35", 
and  its  supplement  152°  51'  25",  which  has  the  same  sign,  and  these  arcs  in- 
creased by  every  multiple  of  27r. 

9.  Find  sin-\62583;  cos-\34268;  tan-M:68531;  cot-\876434. 

Results.     Sin--^62583  =  38°  44'  35",  and  141°  15'  25";   and  these 

arcs  increased  by  every  multiple  of  2'7r. 
cos-\34268  =  69°  57'  36",  and  360°  -  69°  57'  36"  = 

290°  2'  24",  and   these   arcs  increased   by   every 

multiple  of  27r. 
tan~'.468531  =  25°  6'  16",  and  180°  +  25°  6'  16"  =  205° 

06'  16",  and  these  arcs  increased  by  every  multiple 

of  2;r. 
cot-\876434  =  48°  46'  3",  and  180°  +  48°  46'  3"  = 

228°  46'  03",  and  these  arcs  increased  by  every 

multiple  of  27r. 


CONSTRUCTION  AND  USE  OF  TEIGONOMETRICAL  TABUIS.  37 

SuG. — Observe  that  an  increase  of  the  arc  makes  a  decrease  of  its  co-functions. 
In  the  table  of  tangents  as  given,  Table  IV.,  the  proportional  parts  given  at  the 
bottom  of  each  column  are  the  approximate  changes  which  the  functions 
undergo  for  a  change  of  1"  in  the  function.  Thus,  in  finding  cot-*.87G434,  we 
find  cot-*  .876462  =  48°  46' ;  and  at  the  bottom  we  find  that  a  change  of  8.64 
(million ths)  in  the  function  makes  a  change  of  1"  in  the  angle.  Hence,  as  the 
given  cotangent  is  28  (millionths)  less  than  the  cotangent  of  48°  46',  the  angle 
required  is  28  -;-  8.64  =  3  (seconds),  greater  than  48°  46'. 

71»  Sen. — It  is  usually  best  lo  take  from  the  table  that  function  which  is  nearest 
in  value  to  the  given  function,  and  then  increase  or  diminish  the  corresponding 
arc  as  the  case  may  require.  If  wx  always  take  from  the  table  the  next  less 
function  than  that  given  in  the  example  for  sine,  tangent,  and  secant,  and  the 
next  greater  for  the  cosine,  cotangent,  and  cosecant,  coiTections  for  seconds 
will  require  alwaj^s  to  be  added.  If  we  always  take  from  the  tables  the  func- 
tions next  less  than  the  one  given,  the  corrections  for  seconds  must  be  added  for 
sine,  tangent,  and  secant,  and  subtracted  for  the  co-functions.  If  we  were  always 
to  take  from  the  tables  the  next  greater  function  than  the  one  given,  the 
seconds  corrections  would  be  added  for  the  co-functions,  and  subtracted  for  the 
others. 

[Note. — It  is  very  important  that  the  pupil  become  so  familiar  with  the 
nature  of  these  tables  as  to  use  them  intelligently,  and  not  mechanically.  For 
this  reason  we  refrain  from  giving  the  usual  specific,  mechanical  directions  for 
their  use,  and  substitute  illustrations  showing  how  they  are  used  in  accordance 
with  the  principles  upon  which  they  are  constructed.] 

10.  Find  sin-^- .34256);  cos-^- .62584);  tan-^(-- 3.41621) ; 
cot-^(-  1.21648). 

Results.    sm-X-.34256):=200°l'58",and339°58'02",  and  these 

arcs  increased  by  every  multiple  of  2ir. 
cos-'(-.62584)  =  128°  44'  38",  and  231°  15' 22",  and  these 

arcs  increased  by  every  multiple  of  2-^'. 
tan-H-  3.41621)  =  106°  18'  57",  and  286°  18'  57",  and 

these  arcs  increased  by  every  multiple  of  2t. 
cot-^-  1.21648)  =  140°  34'  42",  and  320°  34'  42",  and 

these  arcs  increased  by  every  multiple  of  2-^. 

8uG. — To  obtain  these  results  the  pupil  will  need  to  recall  the  principles  m 
the  corollaries  to  {4S—55).  Thus,  to  find  cot- \  -  1.21648),  we  find  from 
the  table  that  cot-'(1.21648)  =  39°  25'  18";  and  from  (55)  Cor.,  we  learn  that 
cot  (180°-  X) :-  -  cot  X.  .-.  Cot-»(-  1.21648)  =  180°  -  39°  25'  18"  =  140°  34'  42". 
Again,  from  the  same  corollary,  we  learn  that  cot  (360°  —  x)  =  —  cot  x. 
\  Cot-'(-  1.21468)  =  360°  -  39°  25'  18"  =  320°  34'  42". 


11.  Gi-^en  the  logarithmic  sine  9.451234,  to  find  the  corresponding 
Angle. 


38  PLAN'S  TEIGONOMETRY. 

Solution. — The  next  nearest  log  sin  found  in  Table  II.,  is  9.451204  =  log 
sin  16°  2o'.  Now  we  learn  from  the  table  that  an  increase  of  1"  in  the  angle  at 
this  point,  makes  an  increase  in  its  log  sin  of  7.14  (millionths).  But  the  given 
log  sin,  9.451234,  is  30  (millionths)  greater  than  log  sin  16°  25'.  /.  The  required 
angle  is  30  h-  7.14  =  4  (seconds)  greater  than  16°  25' ;  and  we  have  sin  16°  25'  4" 
=  0.451234.  Again,  as  sin  16°  25'  4"  =  sin  (180°  -  16°  25'  4")  =  sin  163°  3^'  50 
the  latter  angle  has  for  its  log  sin  9.451234.  Finally,  either  of  these  augles  in, 
creased  by  any  multiple  of  2Tt  has  the  same  logarithmic  sine. 

12.  Show  from  the  table  that  the  angle  whose  log  cos  is  9.778151, 
is  53°  7'  49",  and  also  306°  52'  11",  and  each  of  these  angles  increasec^ 
by  any  multiple  of  '2ir. 

13.  What  angles  correspond  to  the  logarithmic  cosines  9.246831. 
and  9.889372? 

14.  Find  from  the  table  what  angles  have  for  their  logarithmic 
tangents  9.895760,  10.531054,  and  11.216313. 

Results.  The  first  two  are  the  log  tans  of  38°  11'  20",  and  73° 
35'  43",  and  also  of  180°  +  either  of  these  angles,  and  each  increased 
by  any  multiple  of  2t, 

15.  Find  the  angles  corresponding  to  the  logarithmic  cotangents 
10.008688,  9.638336,  and  9.436811. 

Results,  The  first  two  are  the  log  cots  of  44°  25'  37",  and  66°  29' 
54",  and  also  of  180°  +  either  of  these  angles,  and  each  increased  by 
any  multiple  of  2-^. 


72,  ScH.— Strictly  speaking,  negative  numbers  have  no  logarithms;  since 
no  base  can  be  assumed,  such  that  all  negative  numbci-s  can  be  represented  by 
said  base  affected  with  exponents.  It  is  therefore  customary  to  say  that  nega- 
tive numbers  have  no  logarithms.  Nevertheless,  ue  do  apply  logarithms  to  nega- 
tite  trigonometriMl  functions.  Thus,  if  we  have  —  cos  r,  the  —  sign  is  inter- 
preted as  simply  telling  in  what  quadrants  x  may  end;  while,  in  other  respects, 
the  function  is  treated  exactly  like  +  cos  x. 

16.  Given  log  (-  cos  a;)  =  9.346251,  to  find  x. 

Solution. — The  logarithmic  cosine  9.346261,  considered  independently  of  its 
sign,  corresponds  to  77"  10'  35".  But  the  —  sign  requues  that  the  arc  shall  end 
in  the  2d  or  3d  quadrant,  for  such  angles,  and  such  only,  have  negative  cosines. 
•.  The  angles  required  are  180°  T  77°  10'  35"  =  102°  49'  25".  and  257°  10'  35", 
and  these  increased  by  entu-e  circumferences,  as  all  these  angles  have  loga- 
rithmic cosines,  which  are  numerically  equal  to  9.3462G1,  and  the  cosines  them- 
selves are  negative. 

17.  What  angle  less  than  180°  has  a  negative  cosine  whose  tabu 
lar  logarithmic  value  is  9.653825  ?  Ans.  116°  47'  4", 


CO^■STEUCTION  AND   USE   OF   TEIGONOMETRICAL  TABLES.  o9 

18.  What  angle  lebj  than  180°  has  a  negative  tangent  whose  tabu 
lav  logarithmic  value  is  9.884130  ?  A?is.  142°  33'  15" 

19.  What  angle  less  than  180°  has  a  negative  sine  whose  tabular 
logarithmic  value  is  9.341627  ? 

20.  What  angle  less  than  180°  has  a  negative  cotangent  whose 
tabular  logarithmic  value  is  9.564299  ?  A7is.  110°  8'  15". 

21.  Find  values  of  ic  <  180°  which  fulfil  the  following  conditions : 

log  (-  cos  x)  =  9.562468 ;  log  (-  tan  x)  =  10.764215 ; 
log  (-  sin  x)  =  8.886432;  log  (-  cot  x)  =  11.152161. 

Eesults,  111°  25';  99°  45'  54";  none;  175°  58'  14". 

22.  Having  at  hand  only  the  common  logarithmic  tables  of  trig- 
onometrical functions,  and  the  table  of  logari thins  of  numbers,  I 
wish  to  find  the  number  of  degrees,  minutes,  and  seconds  corre- 
sponding to  the  natural  tangent  2.16145.  How  is  it  done,  and  what 
is  the  result? 

Ansiuer:  Find  the  logarithm  of  2.16145,  to  this  add  10,  and  fii^d 
the  angle  corresponding  to  this  tabular  logarithmic  tangent.  The 
angle  is  65°  10'  20". 

23.  From  the  same  tables  as  above  find  the  natural  cosine  of 
35°  23'.    Also  what  angle  corresponds  to  natural  tangent  2. 

24.  From  the  same  tables  as  above  find  the  angle  corresponding  to 
natural  tangent  —  1.82645.    Also  to  natural  cosine  —  .42536. 

25.  Why  is  it  in  the  table  of  logarithmic  functions  that  the  sine 
of  an  angle  minus  its  cosine  +  10  gives  the  tangent  ?  Why  that  cosine 
—  the  sine  +  10  gives  the  cotangent  ?  Wliy  that  the  sum  of  the  tan- 
gent and  cotangent  of  any  angle  =  20  ?  Why  is  but  one  column  of 
tabular  differences  needed  for  tangents  and  cotangents,  while  the 
sines  and  cosines  require  each  a  separate  column  ? 


(7] 


FUISCTIONS  OF  ANGLES  NEAR  THE  LliMITS  OF  THE  QUADUANT. 

TABLE  III. 

[Note. — This  subject  may  be  omitted  in  an  elementary^ course,  the  first  time 
going  over,  if  thought  best.] 

73,  Failure  of  Table  II. — The  method  which  has  been  given 
in  the  preceding  pages  for  finding  the  logarithmic  functions  of  angles 
involving  seconds,  by  means  of  the  Tabular  Differences,  Table  II.,  in 
Bufliciently  accurate  in  most  cases   for  practical  purposes,  but  ia 


40  PIANE  TRIGONOMETRY. 

entirely  too  rude  for  the  siues,  tangents,  and  cotangents  of  angles 
near  the  beginning  of  the  quadrant  (those  less  than  2°  or  3°),  and 
for  cosines,  tangents,  and  cotangents  of  angles  near  the  close  of  the 
quadrant  (those  between  87°  or  88°  and  90°).  An  example  will 
render  this  clear.  Suppose  we  wish  to  find  log  sin  1'  12".  We  find 
from  Table  II.,  log  sin  1'  =  6.463726 ;  and  also  that  the  average 
increase  of  the  log  sin  between  1'  and  2'  is  5017.17  (million tLs)  for 
every  second  increase  of  the  angle.  But  this  average  rate  of  increase 
©f  the  function  dui-ing  the  minute  is  much  less  than  its  real  rat^  of 
increase  ^;^  the  first  part  of  the  minute,  as  from  I'to  1'12",  and  much 
greater  than  the  real  rate  of  increase  in  the  latter  part.,  as  the  angle 
approaches  2'.  In  fact,  we  see  from  this  table,  that  we  should  use 
2934.85  as  the  increase  of  log  sin  2'  for  1"  increase  of  the  arc.  Now, 
in  our  proposed  example,  we  want  the  increase  of  the  log  sin  while 
the  angle  is  passing  from  1'  to  1'  12".  This,  as  shown  above,  is  con- 
siderably more  than  5017.17  (millionths)  for  every  second. 

TJie  cosine  being  the  sine  of  the  complement  is  subject  to  the  same 
law  of  change  near  the  close  of  the  quadrant,  that  governs  the  sine 
at  the  beginning. 

The  case  of  the  tangent  of  a  small  angle  is  similar  to  that  of  the 
sine ;  and  since  the  cotangent  is  the  reciprocal  of  the  tangent,  it 
has  the  same  laiu  of  change,  only  that  the  one  increases  as  the  other 
decreases.  Thus,  since  doubling  a  small  arc,  as  1",  doubles  its  tan- 
gent (approximately),  it  divides  its  cotangent  by  2. 

Finally,  while  the  law  of  change  in  the  sine  is  very  different  neai 
the  close  of  the  quadrant  from  what  it  is  near  the  beginning,  the 
sine  changing  very  rapidly  at  the  beginning  and  very  slowly  at  the 
close,  and  the  cosine  is  just  the  opposite,  the  tangent,  and  cotangent 
have  the  same  law  of  change  at  both  extremities  of  the  quadrant. 
Thus,  if  near  the  beginning  of  the  quadrant  a  certain  small  increase 
of  the  arc  increases  the  tangent  at  a  particular  rate,  it  decreases  the 
cotangent  at  the  same  ra/e,- since  these  functions  are  reciprocals  of 
each  other.  Moreover,  since  tan  1°  =  cot  (90°  —  1°)  =  cot  89°, 
cot  89°  changes  according  to  the  same  law  as  tanl°;  and  tan  89 
changes  reciprocally  with  cot  89°. 

74.  JDescrijytion  of  Table  III, — The  first  page  of  the  table 
enables  us  to  find  the  sines  of  angles  less  than  2°  36'  15"  (and  con- 
sequently the  cosines  of  angles  between  87°  23'  45"  and  90°)  with 
Very  great  accuracy.  The  columns  headed  Angles  contain  the  degrees, 
minutes,  and  seconds  of  the  proposed  angles,  and  the  columns  at 
their  right  give  the  same  angles  in  seconds.    The  columns  headed 


CONSTEUCTION  AND  USE  OF  TEIGONOMETRICAL  TABLES.    41 

Diff.  contain  tlie  corrections  to  be  used  according  to  th6  following 
problems.  The  second  and  third  pages  answer  a  similar  purpose 
with  reference  to  tangents  and  cotangents  of  arcs  within  2°  36'  20" 
of  the  limits  of  the  quadrant. 


75*  ^rop. — Letting  x  represent  any  number  of  seconds  less  than 
2°  36'  15",  lue  have, 

log  sin  x"  =  4.685575  +  logx  —  Diff. 

Dem.— The  length  of  1"  of  an  arc  to  raduis  unity  is  3.14159265358979  (the 
length  of  the  semicircumference)  -f-  648000  (the  number  of  seconds  in  180"),  and 
=  .00000484812.  FOr  practical  purposes  this  fraction  may  also  be  taken  as  the 
sine  of  1".  Though,  actually,  the  sine  is  less  than  the  arc,  the  expressions  for 
arc  1"  and  sine  1"  agree  to  as  many  places  of  decimals  as  we  have  here.  Again, 
for  these  small  arcs  the  sine  increases  at  nearly  the  same  rate  as  the  arc,  so  that 
sin 3"  =  .00000484812  x  3  nearly;  sin  102"  =  .00000484812  x  102  nearly;  these 
results  being  slightly  in  excess  of  the  true  values.  It  is  the  correction  for  this 
excess  that  is  furnished  by  Table  III.  in  the  columns  marked  Diff.  But  this 
table  is  adapted  to  logarithmic  computation;  hence  we  have  log  sin  a;"  =  log 
sin  1"  +  loga;  —  Difif.  In  this  expression  log  sin  x"  is  the  logarithm  of  the  natu- 
ral sine  of  x"  (not  increased  by  10,  as  each  function  in  Table  II.  is) ;  log  sin  1" 
+  logic,  that  is,  the  logarithmic  sine  of  1"  plus  the  logarithm  of  the  number  of 
seconds,  corresponds  to  multiplying  the  sin  1"  by  the  number  of  seconds,  and 
gives  the  Icgarithm  of  the  product,  or  strictly,  the  logarithm  of  the  length  of  the 
arc  of  re".  Now,  the  sine  o(x"  being  less  than  the  arc,  its  logarithm  is  less  than 
the  logarithm  of  the  length  of  the  arc.  Just  how  much  less  the  table  tells.  This 
difference,  therefore,  between  the  logarithm  of  the  arc  and  the  logarithm  of  its 
sine,  which  is  given  in  the  table,  is -to  be  subtracted.  Finally,  to  make  this 
result  agree  with  Table  II.  we  must  add  10  to  the  result.  Now,  log  sin  1"  = 
log  .00000484812  =  6.685575,  and  adding  10,  we  have  4685575. 

.-.  log  sin  a;"  =  4685575  +  \ogx  —  Diff., 
a  result  which  agrees  with  the  logarithmic  functions  in  Table  II.    q.  e.  d. 

76.  Cor.  l.—  To  obtain  the  log  cos  of  an  angle  between  87°  23' 45' 
•%nd  90'',  from  this  table,  take  the  log  sin  of  its  coinplement. 

77.  Cor.  2.  — To  obtain  the  log  tan  of  an  angle  less  than  2°  36'  20", 
from  this  table,  use  the  formula, 

log  tan  x"  =  4.685575  +  logrc  +  Diff. 

Dem.— For  as  small  an  arc  as  1",  sine,  arc,  and  tangent  are  practically 
equal ;  hence,  log  tan  1"  =  log  sin  1''  =  4685575  (10  being  added).  Moreover, 
for  these  small  arcs  the  tangents  increase  (like  the  sines)  in  nearly  the  same 
ratio  as  the  arcs;. hence,  we  add  log  a;.  Finally,  the  tangent  is  a  little  in  excesg 
jf  the  arc,  which  excess  is  given  in  the  table,  and  is  to  be  added,    q.  e.  d. 


^  PLANE  TRIGONOMETRY. 

78.  Cor.  3.— To  oUain  the  log  cot  of  an  angle  less  than  2°  36'  15', 
frojn  this  table,  use  the  formula, 

log  cot  a;"  =  15.314425  -  logo;  -  Diff. 

DEMONSTRATioN.-Since  cot  X"  =  ^^^'  log  cot  X"  =  log  1  -  log  tan  x" 
=  20  -  (4.685575  +  log  x  +  Diff.)  =  15.314425  -\ogx-  Diflf.  The  20  arises 
from  adding  10  twice  to  log  1  (=  0).  One  10  is  added  because  4.685575  is  10  in 
excess  of  the  true  log  tan  1"  ;  aud  the  other  10  is  added  in  order  to  make  the 
log  cot  x"  agree  with  the  ordmary  tabulated  value,  as  in  Table  II.    Q.  e.  d. 


79,  Cor.  4.— To  obtain  the  log  tan  of  an  angle  between  87°  23'  45" 
and  90°,  take  the  log  cot  of  its  C07npleme7it ;  and  to  obtain  the  log  coi 
of  an  angle  between  the  same  values,  take  the  log  tan  of  its  complement. 


80.  JProb. — Having  given  a  log  sin  less  than  8.657397  {the  log 
t>i7i  2°  36'  15''),  to  find  the  corresponding  angle. 

Solution.— From  log  sin  x"  —  4.685575  +  log  x  —  Diflf.,  we  have,  log  x  = 
log  sin  7^'  —  4.685575  +  Diff.  Hence,  if  from  the  given  log  sin,  we  subtract 
4.685575,  and  then  add  the  proper  correction  as  furnished  by  Table  III.,  we  have 
the  logarithm  of  the  number  of  seconds  sought.  But  we  cannot  tell  what  Diff. 
to  take  till  we  know  the  number  of  seconds.  To  meet  this  difficulty,  find  the 
angle  corresponding  to  the  given  log  sin  from  Table  II,,  and  reduce  it  to  seconds. 
This  will  be  sufficiently  accui-ate  to  furnish  the  required  Diff. 


81,  Cor.  1. — Having  given  a  log  cos  less  than  8.657397  {the  log  cos 
of  87°  23'  45"),  to  find  the  corresjjonding  angle,  treat  it  as  if  it  ivere  a 
log  sin,  and  having  found  the  corresponding  angle,  take  its  compile- 
rneJit. 


82,  Cor.  2. — For  log  tan  aiid  log  cot,  the  formulce  in  {77,  78), 
give, 

logx  =  log  tan  x"  -  4.685575  -  Diffl, 

and,log:c  =  15.314425  —  log  cot  a;"  —  Diflf. 

These  are  applied  as  iu  {80) ;  tliat  is,  the  Diflf.  to  be  subtracted  13 
found  by  getting  from  Table  II.  the  required  angle  in  seconds,  a3 
near  as  may  be,  and  then  take  from  Table  III.  the  corresponding 
Diff. 


CONSTRUCTION  AND  USE  OF  TRIGONOMETRICAL  TABLES.  43 

EXAMPLES. 

1.  Fiud  the  log  sin,  tan,  and  cot  of  1°  11'  15". 

Solution.— Log  sin  1°  11'  15"  =  4.685575  +  log  4275  -  .000031  =  8.316480. 
1°  11'  15"  =  4275".  Since  4275"  is  between  4230"  and  4300,  the  Diff.  is  31 
(millionths). 

Log  tan  1°  11'  15"  =    4.685575  +  log  4275  +  .000062  =   8.316573. 
Log  cot  1°  ir  15"  =  15.314425  -  log 4275  -  .000062  =  11.683427. 
Or,  log  cot  can  be  found  by  subtracting  log  tan  fi*om  20. 

2.  Verify  the  following  by  using  Table  III. : 

log  sin  56' 26"  =  8.215242;  log  tan  56' 26"  -   8.215301; 
logcot56' 26"  =11.784699. 

3.  Verify  the  following  by  using  Table  III. : 

log  cos  88°  17'  44"  =  8.473396 ;  log  tan  88°  17'  44"  =  11.526412  ; 
■     loa  cot  88°  17'  44"  =  8.473588. 


4.  Having  given  the  logarithmic  sine  7.246481  to  find  the  angle. 

Solution.— From  Table  XL  we  find  6'  5"  —  3G5"  as  the  angle.  But  this  is 
subject  to  the  inaccuracy  exhibited  in  {73).  To  obtain  the  coiTect  result  from 
Table  III.,  we  have  {80\ 

log  X  =  7.246481  -  4.685575  +  0  =  2.560906. 
.-.  X  =  363.8;  or  the  angle  is  6'  3"  .8. 

5.  Given  the  logarithmic  tangent  7.805487,  to  find  the  correspond- 
ing angle. 

Solution.— Table  IL  gives  21'  58".3  =  1318".3  as  the  angle.  From  Table  III. 
the  DLff.  corresponding  to  this  is  6  (millionths).    Hence, 

log  X  =  log  tan  a;"  -  4.685575  -  Ditf.  {82) 
becomes,  log  x  =  7.805487   -  4.685575  -  .000006  =  3.119906. 
.-.  X  =  1318 ;  and  the  angle  is  1318"  =  21'  58". 

6.  Given  the  logarithmic  cotangent  12.197148,  to  find  the  corre- 
sponding arc. 

Table  II.  gives  the  angle  21'  49".8,  but  the  true  angle  as  given  by 
Table  III.  is  21'  50". 


M 


PLA^•E  TRIGONOMETRY. 


SECTION  V, 

TRIGOOMETRICAL   SOLUTION    OF   PLAXE   TRIANGLES. 

83,  There  are  six  parts  in  every  plane  triangle :  three  sides  and 
three  angles ;  one  side  and  any  other  two  of  which  being  given,  the 
remaining  parts  can  be  found  by  means  of  the  relations  which  exist 
between  the  sides  and  tabulated  trigonometrical  functions.  To 
exhibit  these  highly  important  practical  operations  is  the  object  of 
this  section.  We  shall  treat  first  of  right  angled  plane  triangles,  and 
then  of  oblique  angled  plane  triangles. 

OF  RIGHT  ANGLED  TRIANGLES. 


JProj}.  84, — Tlie  relations  between  the  sides  and  the  trigoiimnet- 
rical  functions  of  the  oblique  angles  of  a  right  angled  triangle  are  as 
foUoios : 


side  opposite  ^ 
hypotenuse   ' 
side  adjacent  ^ 
hypotenuse  ' 
/ox  |.  f  _  side  opposite  ^ 

^  '^    '    °        ~  side  adjacent' 


(1)  sine         = 

(2)  cosine     = 


.*.  cosecant   = 
.*.  secant        = 


cotangent  = 


hypotenuse 
side  opposite ' 
hypotenuse 
side  adjacent' 
side  adjacent 
side  oj^posite* 


FiQ.  12. 


«EM.— Let  CAB,  Fig.  12,  be  a  tiiangle,  right 
angled  at  A.  Let  aM  be  the  mcasuriug  arc  of  the 
angle  B,  PD  =  sin  B,  and  BD  =  cos  B.    From  the 


PD 

similar  triangles  PDB  and  CAB,  we  have  gp  = 

since   BP  =  1. 


CA 
BC 


I.  e. 


.    r>      Side  opposite 

sin  B  = i-* , 

hypotenuse 


From  the  same  tiiangles 


BD 
BP 


BA 
BC 


I.  «.,  cos  B  = 


side  adjacent 


.  Tangent  being  equal  to  sine  divided 


by  cosine,  we  nave  tan  B  = 


hypotenuse 
side  opposite  ^  side  adjacent  _8ide  opposite 
hypotenuse 


The 


hypotenuse  hypotenuse  side  adjacent 
other  functions  being  the  reciprocals  of  these  three,  are  as  given  in  the  proposi- 
tion. Finally,  as  a  similar  construction  could  be  made  about  the  other  obhque 
angle,  C,  this  demonstration  may  be  considered  general,    q.  k.  d. 


SOLUTION  OF  EIGHT  ANGLED  PLANE  TRL^NGLES.  45 

ScH.  1. — These  formula  are  so  important  that  it  is  well  to  have  them  fixed 

in  the  memory,  not  oaly  as  written  above,  but  also  as  follows: 

r,. ,                ,           .              ^J                       .         ,          side-adj 
(IV  Side-opp  =:  hy  x  sm,  or ,  or  tan  x  side-adj,  or ; 

COScC  COL 

mN    a- 1      T      1  ^iv  ^        .,  side-opp 

(2).  Side-adj  =  hy  x  cos,   or  — ^,  or  cot  x  side-opp,  or -^  ; 

-'         -^  sec  ^  tan      ' 

of  which  the  relations  side-opp  =  hy  x  sin,  and  side-adj  =  hy  x  cos  are  of  the 

most  frequent  use. 

ScH.  2. — The  six  ratios  given  in  this  proposition  are  frequently  made  the  def- 
initions of  the  tiigonometrical  functions.  Thus,  referring  to  a  right  angled  tri- 
angle, a  sine  of  an  angle  may  be  defined  to  be  the  ratio  of  the  side  opposite  to 
the  hypotenuse ;  the  cosine  as  the  ratio  of  the  side  adjacent  to  the  hypotenuse, 
etc. 

Sen.  3. — The  student  will  be  aided  in  remembering  these  important  relations 
by  observing  that  the  side  opposite  the  angle  is  analogous  to  the  sine,  and  the 

.,       ,.                  ,                     -..-r         ,       .          its  analogue 
side  adjacent  to  the  cosine.    Now,  the  sine  = r ,  and  so  also  the  co- 
sine.   Tangent  equals  sine  divided  by  cosine,  and  in  this  case  it  is  the  part 
analogous  to  the  sine,  divided  by  the  part  analogous  to  the  cosine.     One  shouM 

hv 

not  make  the  blunder  of  saying  that  sin  =    .  -,   ^ ,  since  tliat  would  make 

side-opp 

tbe  sine  always  more  than  1 ;  but  we  have  seen  that  it  never  can  exceed  1. 

Similar  checks  against  error  may  be  made  in  the  case  of  the  other  relations. 


EXERCISES. 

[Note. — The  first  five  of  these  exercises  are  mainly  designed  to  illustrate  the 
proposition,  and  familiarize  the  mind  with  the  relations.] 

1.  In  a  right  angled  triangle  whose  sides  are  3,  4,  and  5,  what  are 
the  trigonometrical  functions  of  the  angles  ?  What  are  the  functions 
when  the  sides  are  6,  8,  and  10  ? 

2.  In  a  right  angled  triangle  the  hypotenuse  is  12,  and  the  angl^ 
at  the  base  sin~^ J.  What  are  the  sides  ?  What  is  the  sine  of  th/» 
other  angle  ? 


Suo.— Represent  the  angles  by  B,  A,  and  C,  A  being  the  right  angle ;  and 
the  sides  opposite  by  5,  a,  and  c.  Then  sin  B  =  -,  or  ^  =  — ;  whence,  h  =  ^ 
Sin  C  =  cos  B  =  ^y^  c  =  6 yT 

3.  In  a  right  angled  triangle  whose  hypotenuse  is  12,  and  the 
angle  at  the  base  tan~^2,  what  are  the  other  parts  ? 

A71S.  Cos-^f\/5;  ^Vo7and^^/5, 


46  PLANE  TRIGONOMETEY. 

4.  The  sides  of  a  right  angled  triangle  are  .20  antl  32.  What  are 
the  angles  and  hypotenuse?  Obtain  the  hypotenuse  by  means  of 
the  secant.  _ 

Ans.  Tan-4,tan-^f,  and  WS9. 

5.  The  hypotenuse  of  a  right  angled  triangle  is  120,  and  one 
side  100.  Show  that  the  angle  opposite  the  latter  is  tau~^^jVTi,  the 
adjacent  angle  cosec~^3^A/llj  aiid  the  remaining  side  20\/ll.  Obtain 
these  results  in  the  order  given.  Use  a  trigonometrical  function  to 
obtain  the  last. 


EXAMPLES. 

(a)    BY   MEANS  OF  THE   TABLE   OF   I^ATURAL   PUXCTIO:S"S. 

1.  In  a  right  angled  triangle  ABC,  the  hypotenuse  BC  is  235,  and 
the  angle  B  is  43°  25'.     Find  the  angle  C,  and  the  sides  AB  and  AC. 

C  =  46°"35';  AB  =  170.7;  AC  =  161.52. 

SoLunox — To  find  C,  we  have  but  to  remember  that  the  angles  of  a  right 
angled  triangle  are  complements  of  each  other ;  whence,  C  =  90"  —  B  =  46'  35'. 
AR 

To  find  AB,  we  have  cos  B  =  ^'  or  AB  =  235  x  cos  43°  25'.  Now,  from  the  table 
of  natural  functions  we  find  cos  43°  25'  =  .72637;  whence   AB  =  235  x  .72637 

-=  170.7.     To  find  AC,  we  have  sin  B  =  || ;  whence  AC  =  235  x  .6873  =  161.52. 

2.  In  a  right  angled  triangle  ABC,  the  hypotenuse  AC  is  94.6,  and 
the  angle  c  is  56°  30'.    Find  the  angle  A,  and  the  sides  AB  and  BC 

A  =33°  30';  AB  =78.88  ;  BC  =  52.21. 

3.  In  a  right  angled  triangle  BDF,  the  hypotenuse  BF  is  127.9,  and 
the  angle  B  is  40°  10'  30".  Find  tlie  angle  F,  and  the  sides  BD  and 
DF. 

F  =  49°  49'  30";  BD  =  97.72;  DF  =  82.51. 

4.  In  the  triangle  CDE,  right  angled  at  E,  given  the  side  DE  75,  the 
Bide  CE  50.59,  to  find  the  other  parts. 

Hyi^otenuse  =  90.47. 

5.  In  the  right  angled  triangle  CDE,  given  the  hypotenuse  CD  264, 
the  side  CE  135.97,  to  find  the  other  parts. 

DE  =  226.28. 

6.  Given  the  hypotenuse  435,  and  one  of  the  acute  angles  44°,  to 
find  the  other  parts. 

7.  Given  the  hypotenuse  64,  and  the  base  51.778,  to  find  the  other 
parts. 


SOLUTION  OF  KIGHT  ANGLED  PLANE  TRL^NGLES.  47 

8.  GiTeu  the  hypotenuse  749  feet,  and  the  base  548.255  feet,  to 
find  the  other  parts. 

9.  Given  the  hypotenuse  125.7  yards,  and  one  of  the  acute  angles 
75°  12  23",  to  find  the  other  parts. 

10.  Given  one  side  388.875,  and  the  adjacent  angle  27°  38'  50",  to 
find  the  other  parts  of  a  right  angled  triangle. 


(b)    BY   MEAXS   OF  A  TABLE   OF   LOGARITHMIC  FUN"CTIOj;rS. 

11.  In  a  right  angled  triangle,  given  an  oblique  angle  54°  27'  39", 
find  the  side  opposite  56.293,  to  find  the  other  parts. 

Solution.— The  other  oblique  angle  is  90°  -  54°  27'  39"  =  35°  32'  21". 
To  find  (lie  hypotenuse,  we  have  sin  54°  27'  39"  ——j- — ,  or  liy  = 


hy  '  '  sino4°27'89" 
Applying  logarithms  to  facilitate  computation,  log  hy  =  log  56.293  —  log 
sin  54°  27'  39"  +  10.  The  10  is  added  since  the  log  sin  54°  27'  39"  found  in  the 
t<ible  is  10  too  great.  Now,  looking  in  Table  L,  we  find  log  56.293  =  1.750454 ; 
and  in  Table  II.,  log  sin  54°  27'  39"  =  9.910473.    Hence, 

log  7iy  =  1.750454  -  9.910474  +  10  =  1.839980,  and  hy  =  69.18. 

sids  OUT) 

To  find  the  other  side  we  have,  tan  angle  =  — r-; — -jr,  or  tan  54°  27'  39"  — 
•^  '  °  side  adj 

56  293  56  293 

whence  side  adj  = ^    '  .     Applying  logarithms,  log  side  adj 


side  adj  '  "^      tan  54°  27'  39' 

=  log  56.293  -  log  tan  54°  27'  39"  +  10  =  1.750454  -  10.146104  +  10  =  1.604350, 

.-.  Side  adj  =  40.2115. 

12.  In  a  right  angled  triangle  ABC,  the  hypotenuse  AC  is  340,  and 
the  side  AB  is  200.  Find  the  acute  angles  A  and  C,  and  the  perpen- 
dicular BC. 

A  =  53°  58'  6"  ;  C  =  36°  1'  54'  :  BC  =  274  95. 

13.  In  a  right  angled  triangle  ABC,  the  perpendicular  AB  is  736.3 
and  BC  500.    Find  the  acute  angles  A  and  C,  and  the  hypotenuse  AC, 

A  =  34°  10'  45"  ;  c  =  55°  49'  15" ;  AC  =  890.02. 

14.  In  a  right  angled  triangle  BDF,  the  perpendicular  BD  is  246.32, 
and  DF  380.07.  Find  the  acute  angles  E  and  F,  and  the  hypote- 
nuse BF. 

B  =  57°  3'  11";  F  =  32°  56'  49" ;  BF  =  452.91. 

15.  In  a  right  angled  triangle  ABC,  the  side  AB  is  249,  and  the 
angle  A  is  29°  14'.  Find  the  perpendicular  BC,  and  the  hypotenuse 
AC 

BC  =  139.35  ;  AC  =  285.341. 


iS  PLANE  TEIGONOMETET. 

16.  In  a  right  angled  triangle  ABC,  the  hypotenuse  AC  is  1)5.75,  and 

the  side  BC  60.    Find  the  acute  angles  A  and  c,  and  the  pei-pen- 

dicular  AB. 

A  =  38°  48'  r';  C=  51°  11'  53"  ;  AB  =  74.62. 

17.  In  a  right  angled  triangle  ABC,  the  side  BC  is  364.3,  and  tlie 
angle  A  is  50°  45'.     Find  the  perpendicular  AB,  and  the  hypotenuse 

AC. 

AB  =  297.645  ;  AC  =  470.433 

[Note. — The  first  ten  examples  may  be  solved  by  logarithms  if  additional 
exercises  are  needed,  or  these  by  means  of  the  natural  functions.  Also  any  one 
of  the  examples  will  afford  several  others  by  giving  and  requiring  different 
parts.  Thus,  from  Ex.  17,  we  can  give  AB  =  297.045,  A  =  50°  45',  and  require 
the  other  parts,  etc.]. 


GEMERAL  APPLICATIONS, 

1.  Find  the  area  of  a  parallelogram  whose  adjacent  sides  are  28 
and  30  feet,  and  the  included  angle  75°. 

SuG.— First  find  the  altiUide. 

2.  A  railroad  track  463  feet  3  inches  in  length  has  a  uniform  grade 
of  3°.     Show  that  the  vertical  rise  is  24  feet  3  inches,  nearly. 

3.  A  railroad  track  makes  a  vertical  rise  of  150  feet,  by  uniform 
grade,  in  3,000  feet  of  track.     TThat  is  the  grade  ? 

4.  Find  the  apothem  and  radius  of  the  circumscribed  circle  of  a 
regular  heptagon  one  of  whose  sides  is  12  feet. 

5.  Find  the  area  of  a  regular  dodecagon  inscribed  in  a  circle  whose 
radius  is  12. 

6.  The  angle  of  elevation  to  the  top  of  a  steeple  is  47°  30',  a© 
measured  from  a  point  in  the  same  horizontal  plane  as  its  base,  and 
at  a  distance  of  200  feet  from  it.    What  is  the  height  of  the  steeple  ? 

Ans.  218.26.  ft. 

7.  A  tower  103  feet  high  throws  a  shadow  51.5  feet  long,  upon  the 
horizontal  plane  of  its  base  :  what  is  the  angle  of  elevation  of  the 
sun? 

8.  The  angle  at  the  vertex  of  a  right  cone  is  52°  23',  and  the 
slant  height  126  feet :  what  is  the  diameter  of  the  base,  and  what 
the  altitude  ? 


SOLUTION   OP   OBLIQUE   ANGLED   TLANE    TRL^NGLES. 


49 


Fi8.  13. 


9.  In  Fig.  13,  letting  EO  rep- 
resent the  earth  and  M  the 
moon,  the  radius  of  the  earth 
EO  =  3956.2,  and  the  angle 
EMO  *  =  57',  required  to  find 
the  distance  OM,  E  being  a 
right  angle. 

The  distance  of  the  moon  from  the  earth,  as  (jiven  ly  this  compu- 
tation, is  238,613  miles, 

10.  In  Fig.  13,  letting  ON  represent  a  tangent  to  the  moon's  disc 
at  N,  the  angle  NOM  is  readily  measured,  being  half  the  moon's 
api)arent  diameter.  The  apparent  diameter  of  the  moon  being 
31'  20",  and  its  distance  from  the  eartli  as  found  in  the  last  example, 
what  is  the  diameter  ? 

Ans.  2176  miles. 


4 


OF  OBLIQUE  ANGLED  PLANE  TRIANGLES. 


IMPORTANT  RELATIONS  EXISTING  BETWEEN  THE  SIDES  AND  TRIGO- 
NOMETRICAL FUNCTIONS  OF  THE  ANGLES  OF  OBLIQUE  ANGLKI) 
PLANE  TRIANGLES. 


85,  I*rop, — Hie  sides  of  any  2)lane 
triangle  are  projyortional  to  the  sines  of 
the  angles  opposite. 

Dem. — Let  ABC  be  any  plane  triangle.  Let 
fall  from  either  angle,  as  C,  a  perpendicular 
upon  the  opposite  side,  or  upon  that  side 
produced.  Designate  the  angles  by  A,  B,  and 
C,  tlie  sides  opposite  by  a,  J,  and  c,  and  the  per- 
pendicular by  P. 

"Now,  from  the  right  angled  triangle  ADC 
we  have  P  =  &  sin  A  ;  also  from  CDB,  P  = 
a  sin  B ;  sin  ABC  in  the  second  figure  being  = 
sinCBD.  Hence,  equating  tlie  values  of  P, 
6  sin  A  =:  «  sin  B,  or  rt  :  6    :  :    sin  A   :  sin  B 


Fig.  14. 


Q.  E.   D. 


*  This  angle  is  called  the  moon's  horizontal  parallax,  and  is  readily  measured.  Some  ruUe 
notion  of  the  manner  in  which  parallax  becomes  apparent,  may  be  pot  from  coni*idering  ihc 
difference  in  direction  from  two  observers  to  the  moon,  one  observer  standing  directly  under 
the  moon,  as  at  F,  and  the  other  at  E,  Beeinij  the  moon  in  his  horizon.  The  an>^iilar  displace 
neut  of  the  moon  due  to  these  different  points  of  observation  is  horizontal  pariillax. 


50 


PLANE  TRIGONOMETEY. 


86,  Pro2)»—TIie  sum  of  any  two  sides  of  aj^^ctne  triangle  is  to 
iJieir  difference,  as  the  tangent  of  half  the  sum  of  the  angles  opposite 
is  to  the  tangent  of  half  their  difference. 

Dext. — Letting  a  and  h  represent  any  two  sides  of  a  plane  triangle,  and  A 
and  B  the  angles  opposite,  we  have  a  :  6  :  :  sin  A  :  sin  B.  Taking  this  both  by 
composition  and  division,  we  have  a  +  b  :  a  —  b  : :  sin  A  +  sin  B  :  sin  A  —  sin  B. 
But  from  (60),  sin  A  +  sin  B  :  sin  A  —  sin  B  : :  tan  i(A  +  B)  :  tan  ^A  —  B). 
.-.   a  +  J  :  a  —  6  :  :  tan  ^A  +  B)  :  tan  \{^  —  B).    q.  e,  d. 


87*  JProp. — Tlie  tangent  of  half  of  any  angle  of  a  plane  triaiigle 
equals  Ic,  divided  hy  half  the  perimeter  of  the  triangle  minus  the  side 
opposite  the  angle  ;  in  which  h  is  the  radius  of  the  inscribed  circle, 
and  equals  the  square  root  of  the  contiiiued  product  of  half  the  peri- 
meter minus  each  side  separately,  divided  hy  half  the  perimeter. 


DEif. — Let  ABC  be  any  plane  ti'iangle. 
Represent  the  angles  by  A,  B,  and  C,  the 
sides  opposite  by  a,  b,  and  c,  the  perimeter 
by  Py  and  the  segments  of  the  sides  made 
by  the  radii  of  the  inscribed  circle,  by  a*,  y, 
and  z,  as  in  the  figure. 

Then  a  +  b  -\-  c  =  %x  +  2y  +  2z  =  p,OT 
u  +  y  +  2  r=  4^. 

Whence  x  =  \p  —  a,  y  =lp  —b,  and  z 


Fro.  15. 


=  iP  —c  ;  since  y  +  z  =  a,x  +  z  =  b,  and  x  +  y  =  c. 


k  k 

Now  from  the   triangles  AOD,  DOB,  and  COE,  tan  ^^=-=-, ,  tan 

X      ip  —  a 

^B  =  ^  =  -A_  and  tan  ^0  =  -  =  ^^—. 
y      ip-b  z      ip-c 

To  find  k.    ^A  +  iB  +  iC  =r  90°,  or  iA  =  90°  -  QB  +  ^C) ;  whence,  tan  i^ 
=  I  =  tan  [90-  -  (iB  +  ^C)]  =  cot  «B  +  iC)  =  j"'^"^^^  (54,  Dem). 


Substituting  for  tan  ^B,  and  tan  ^C,  -,  and  -,  we  have   -  = ^• 

y  s  X       k      k' 


whence 


**  =  ^  ^^-r  •Ax-\-y  +  z)k'  =  xyz ;  and  k  =  |/--^^_.  In  this  value  of  A:,  sub- 
stituting for  a-,  y,  and  z  their  values,  we  have  k  =  j/^^^  ~  ^^  ^^^  ~  ^^  ^^^  ~  ^\ 


SOLUTION  OF  OBLIQUE  ANGLED  PLANE  TRIANGLES.      51 

88,  ScH.  1. — These  three  propositions  {85,  80,  87),  furnish  the  most 
elegant  and  expeditious  means  for  finding  the  unknown  parts  of  an  oblique 
angled  plane  triangle,  when  a  sufficient  number  of  parts  are  given  or  known  (^.5) 

89,  Sen.  2.— Important  Practical  Suggestions, 

1st.  Two  angles  of  a  triangle  being  given,  the  third  is  known  by  implication, 
It  being  the  supplement  of  the  sum  of  the  other  two. 

2nd.  When  two  of  the  known,  or  given,  parts  are  opposite  each  other,  the 
first  proposition  {85)  effects  the  solution. 

3rd.  When  two  sides  and  the  included  angle  are  given,  the  solution  is  effected 
by  means  of  the  second  proposition  {80). 

4th.  When  the  three  sides  are  given,  the  angles  are  found  by  the  third 
proposition  {87). 


EXERCISES. 

1.  In  the  plane  triangle  CDE,  given  the  angle  D  =  15°  19'  51", 
C  =  72°  44'  05'',  and  the  side  c,  opposite  C, 
250.4,  to  find  the  other  parts.  ^ 

d       ^^^^^^ 

Solution.— i^rs^,  E  =  180°-  (D  +  C)  =  91°  56'  04"     L ^^-- 

{89, 1st).  ^ 

Secondy  To  find  side  d  opposite  angle  D  {89^  2nd), 
sin  C  :  sin  D  :  :  c  :  <?,  or 
sm  72°  44'  05"  :  sin  15°  19'  51"  :  :  250.4  :  d. 
This  proposition  may  be  solved  for  d,  by  taking  the  natural  sine  of  15°  19'  51", 
multiplying   it  by  250.4,  and  dividing  the  product  by  the  natural  sine  of 
72°  44'  05";  or,  more  expeditiously,  by  logarithms,  as  follows : 

log  250.4= 2.398634 

log  sin  15°  19'  51"  = 9.422249 

log  sin  72°  44'  05"  (ar.  comp.)*  = 0.020024 

\ogd= t  1.840907    .-.  cf  =  69.328. 

Third,  To  find  side  e  opposite  angle  E  {89^  2nd), 
sin  C  :  sin  E  :  :  c  :  e,  or 
sin  72°  44'  05"  :  sin  91°  56'  04"  :  :  250.4  :  e. 
Making  the  computation  by  logarithms, 

"  log  250.04= 2.398634 

log  sin  91°  56'  04"  % 9.999752 

log  sin  72°  44' 05"  (ar.  comp.)  = 0.020024 

log  e  =    .    .    . 2.418410.    .-.  e  =  262.066. 

*  See  Introduction  to  Table  I.  (17). 

t  The  student  must  bear  in  mind  the  fact  that  all  the  log.  trig,  funcs.  are  10  too  large,  and 
must  pee  exactly  what  corrections  to  make  in  his  results,  on  this  account.  In  this  case  the 
ar.  comp.  is  10  too  small,  since  the  logarithm  we  took  from  the  table  for  log  sin  72°  44'  05"  was 
10  too  large.  But  our  log  sin  15"  19'  51"  is  10  too  large,  and  jast  corrects  the  latter.  Hence,  we 
have  to  reject  only  10  from  the  entire  srm  11.840907,  and  this  on  account  of  the  use  of  ar.  comp. 

t  Take  the  sine  of  the  supplement,  oi  the  cosine  of  the  given  angle  minus  90°. 


52  PLANE  TKIoOXOMETKY. 

2.  lu  the  plane  triangle  ABC,  A  =  35°  42',  B  =  76°  27',  an  J  AB  = 
142.     What  are  the  other  parts  ? 

Ans.  C  =  67°  51';  AC  =  149.05;  BC  =  89.47. 

3.  Given  two  angles  of  a  plane  triangle,  23°  40'  32"  and  69°  39'  51", 
and  the  included  side  100,  to  find  the  other  parts. 

4.  In  a  plane  triangle  ABC,  the  side  AB  is  254.3,  the  side  AC  396.8, 
and  the  angle  B  94°  29'.    Find  the  angles  A  and  C,  and  the  side  BC. 

A  =  45°  48'  21";  C  =  39°  42'  39";  BC  =  285.37. 

Suq's.— To  find  the  angle  C,  we  have 

396.8  :  254.3  : :  sin  94°  29'  :  sin  C. 

From  this  proportion  "we  get  log  sin  C  =  9.805443.  Now,  as  we  have  seen 
before,  there  are  an  infinite  number  of  angles  corresponding  to  any  given  sine, 
how  shall  we  know  what  one  to  take  in  this  case  ?  First,  no  angle  of  a  triangle 
can  exceed  180^ ;  hence,  there  are  but  iico  angles,  one  an  acute  angle,  and  the 
other  its  supplement,  which  can  come  into  consideration  in  the  solution  of  i)lane 
triangles.  But  which  of  these  two  are  we  to  take  ?  Thus,  in  this  case,  both  the 
angles  39°  42'  39"  and  140°  17'  21"  correspond  to  log  sin  9.805443.  In  this  ex- 
ample the  ambiguity  is  resolved  by  observing  that  the  given  angle  B  is  obtuse, 
and  a  plane  triangle  can  have  but  one  obtuse  angle.    .'.  C  =  39"  42'  39". 

5.  In  a  plane  triangle  BDF,  the  angle  B  is  40°,  the  side  BD  is  400, 
and  the  side  DF  350.     Find  the  angles  D  and  F,  and  the  side  BF. 

Sug's.— To  find  F,  we  have 

350  :  400  :  :  sin  40°  :  sin  F, 
from  whioh  log  sin  F  =  9.86G0o9,  and  F  =  47°  16'  28",  and  its  supplement 
132'  43'  32".  How  are  we  to  determine  which  of  these  to  take?  The  given 
angle  is  40° ;  hence,  as  far  as  that  is  concerned,  either  of  the  two  will  meet  the 
condiUons.  There  are,  therefore,  two  angles,  F  =  47°  16'  28"  and  F  =  132'  43'  3.3", 
which  fulfill  the  conditions.  We  therefore  solve  two  triangles,  one  having  two 
of  its  sides  400  and  350,  and  the  angles  40', 
47°  16'  28",  and  92°  43'  32" ;  and  another  Uiaugle 
with  the  same  given  parts  and  the  two  required 
angles  132°  43'  32",  and  7°  16'  28".  This  is  readily 
illustrated  geometrically.  Thus,  lay  otf  DBF'  =  40°. 
Take  BD  =  400.  Then  from  D  as  a  centre,  with 
Fig.  16.  radius  350  describe  an  arc  cutting  BF'.    It  is  evident 

that  if  B  is  an  acute  angle  the  following  cases  may  arise  depending  upon  the 
value  of  DF: 

1st.    If  DF  is  less  than  the  perpendicular  p,  the  problem  is  impossible. 
2nd.  If  DF  =p,  the  triangle  is  right  angled  at  F. 

3rd.  If  DF  >  p  and  <BD  there  are  two  triangles,  one  with  an  acute  angle  at 
F',  as  DF'B,  and  the  other  with  an  obtuse  angle  at  F,  as  DFB,botli  of  which 
fulfil  the  conditions  of  the  problem. 

4th.  If  DF  >  DB  there  is  but  one  triangle  which  fulfills  the  conditions,  viz., 
the  one  with  an  acute  angle  at  F'. 


SOLUTION  OF  OBLIQUE  ANGLED  PLANE  TRIANGLES.      53 

The  results  in  the  above  examples  are,  for  triangle  DF'B,  angle  DFB  .— 
47"  16' 28",  BDF' =  92°  43' 33",  and  side  BF'  =  543.89;  for  triangle  DFB, 
angle  DFB  =  132°  43'  32",  angle  BDF  =  7°  IG'  28",  and  side  BF  =  68.94 

90,  Cor. — Li  applying  trigonometrical  formulce  to  the  solution  of 
triangles,  if  the  part  sought  is  found  in  terms  of  its  sin"e,  the  result 
is  amhiguouSf  and  we  are  to  determine  zuhether  there  really  are  two 
solutions  to  the  prollem  in  a  geometrical  sense,  ty  certain  geometrical 
considerations,  or  else  hy  tryijig  loth  values  for  the  angle  determined 
ly  its  sine.  This  ambiguity  arises  only  tohen  an  angle  is  determined 
hy  its  sine,  as  will  appear  hereafter. 

6.  Given  two  sides  of  a  plane  triangle  201  and  140,  and  the  angle 
opposite  the  latter  36°  44'.     Find  the  other  parts. 

Results. — There  are  two  triangles. 

Parts  of  the  first,  120°  49'  49",  22°  26'  11",  and  89.34; 
Parts  of  the  second,  59°  10'  11",  84°  5'  49",  and  232.84. 

7.  Given  two  sides  of  a  plane  triangle  180,  100,  and  the  angle  op- 
posite the  former  127°  33',  to  find  the  other  parts. 

There  is  but  one  triangle,  and  the  parts  are  26°  7'  59",  2G°  19'  1", 
and  100.65. 

8.  Given  two  sides  of  a  plane  triangle  30.8  and  54.12,  and  the 
angle  opposite  the  latter  36°  42'  11",  to  find  the  other  parts.  Why 
but  one  triangle  ? 

9.  Given  two  sides  of  a  plane  triangle  GOO  and  250,  and  the 
angle  opposite  the  latter  42°  12'.   Find  the  other  parts. 

SuG. — Attempting  to  get  the  angle  opposite  600,  we  find  log  sin  =  10.207400, 
which  is  impossible.  It  is  in  some  such  way  that  a  trigonometrical  solution 
shows  a  geometrical  absurdity. 

10.  Given  two  sides  of  a  plane  triangle  1337.5  and  493,  and  the 
angle  opposite  the  former  69°  46'.  Find  the  other  parts. 


11.  In  a  plane  triangle,  given  two  sides  1686  and  960,  and  the  in- 
cluded angle  128°  04',  to  find  the  other  parts. 

c 

Solution. -Let  a  —  1686,  h  =  960,  and 

C  =  128°  04'.  {See  89,  3rd.)  The  sum  of 
the  angles  A  and  B  is  180°  -  128°  04'  = 
51°  56',  and  i{^  +  B)  =  25°  58'.  From  (86) 
we  have, 

a  +  b  :  a  —b  ::  tan  J,(A  +  B)  :  tan  KA  —  B),  or 
2646  :  726  :  :  tan  25°  58'  :  tan  i{A  -  B). 


54:  PLANE  TRIGONOMETRY. 

Making  the  computations  by  logarithms,  we  find  log  tan  KA  —  B)  =  9.1258S7 
Hence,  ^(A  —  B)  =  7°  36'  40",  the  angle  found  in  the  table,  or  its  supplement 
But  ^  the  difference  of  two  angles  of  a  ti'iangle  is  less  than  90° ;  consequently 
KA  -  B)  =  7=  36'  40". 

Now  having  i(,A  +  B)  =  25°  58',  and  i(A  -  B)  =  7°  36'  40",  we  find  A  = 
33°  34'  40",  and  B  =  18°  21'  20". 

The  side  c  can  be  found  by  (So)  as  two  opposite  parts  are  now  known. 
c  =  2400. 

12.  In  a  plane  triangle  ABC,  the  side  AB  is  304,  BC  280.3,  and  the 
included  angle  B  is  100°.    Find  the  angles  A  and  C,  and  the  side  AC. 

A  =  38°  3'  3" ;  C  =  41°  56'  57"  ;  AC  =  447.856. 

13.  In  a  plane  triangle  ABC,  the  side  AB  is  103,  AC  126,  and  the 
included  angle  A  is  56°  30'.   Find  the  angles  B  and  c,  and  the  side  BC. 

B  =  72°  20'  15" ;  C  =  51°  9'  45"  :  BC  =  110.267. 


14.  In  a  plane  triangle  ABC,  given  the  three  sides,  a  =  3459,  h 
4209,  and  c  =  6030.4,  to  find  the  angles. 


Solution. — Applying  (57),  we  have, 


y  ip  ' 

log^  =  i{log  (ip-a)  +  log  {ip  -b)  +  log  (^  -  c)  -  log ip}. 

k                            k                                   k 
Also,  tan  -iA  = ,  tan  |B  = -,  and  tan  ^C  = , ,  or  log  tan  ^A 

=  log  ^  —  log  {ip  —  a),  log  tim  ^  B  =  log  A;  —  log  (^  —  b),  and  log  tan  ^C  - 
log  k  -  log  {ip  -  c). 

COMPUTATION. 
a  =    3459 
b=    4209 
c=    6030.4 
p  =  13698.4 
ip  =    6849.2     (ar.  comp.)  log  =    6.164360 

ip-  a=    3390.2 log  =    3.530226 

Ip-  b=    2640.2 log  =    3.421637 

ip  -  c  =     818.8 log  =    2.913178 

2)6.029401 
log  k  =    3.014700 
log  tan  ^A  =  log  A;  -  log  {^p  -  a)  =    9.484474    .'.  A  =    33"  56'  10".5 
log  tan  iB  =  log  A;- log  (i;?-  6)=    9.593063    .'.  B  =   42M7'  25'.3 
log  tan  iC  =  log  A;  -  log  {ip  -  c)  =  10.101522    .-.  C  =  103°  16'  24".2 

Proof,  A  +  B  +  C  =  180*  00'  00" 


SOLUTION  OF  OBLIQUE  ANGLED  PLANE  TRIANGLES. 


55 


15.  In  a  plane  triangle  ABC,  the  side  AB  is  95.6,  BC  is  275,  and  AC 
300.    Find  the  angles  A,  B,  and  C. 

A=  65°  47'  55";  B  =  95*»  42'  52";  C  =  18°  29'  13". 

16.  In  a  plane  triangle  BDF,  the  side  BD  is  500,  DF  is  403.7,  and 
BF  395.75.     rind  the  angles  B,  D,  and  F. 

B  =  52°  0'  3";  D  =  50°  34'  45";  F  =  77°  25'  12". 


OBLIQUE  ANGLED  TRIANGLES  SOLVED  BY  MEANS  OF  RIGHT 
ANGLED  TRIANGLES. 

[Note. — Articles  91-94  inclusive,  may  be  omitted  in  an  elementary  course, 
if  thought  desirable.  Or  91  and  its  applications  may  be  taken  instead  of  56*- 
88,  85  should  be  included  in  any  course.  It  is  too  elementary  and  important 
to  be  omitted.] 

91.  I*rop, — All  cases  of  oUique  angled  plane  triangles  may  he 
solved  ly  the  solution  of  right  angled  triangles. 

Dem.— Of  the  three  given  parts  we  may  affirm  that  they  are,  1st,  All 
adjacent;  2nd,  Two  adjacent  and  one  separated;  or  3rd,  All  separated. 

1st.  When  tlie  given  parts  are  all  adjacent ;  i.  d.,  when  they  are  tico  sides  and 
the  included  anrjle,  or  two  angles  and  the  in- 
cluded side.  To  solve  the  first  let  fall  a  perpen- 
dicular from  the  extremity  of  one  of  the  given 
sides  upon  the  other  given  side,  or  upon  that 
side  produced.  There  will  thus  be  formed  two 
right  angled  triangles  which  can  be  computed, 
and  from  the  parts  of  which  the  parts  of  tlie 
required  triangle  can  be  found.  Thus,  let  A  be 
the  given  angle,  and  b  and  c  the  given  sides. 
In  the  right  angled  triangle  ADC  there  are  given 
A  and  5,  whence  AD,  P,  and  angle  ACD,  can  be 
computed.  Then  passing  to  triangle  CDB,  we 
know  P,  and  DB  (since  we  have  c  given  and 
have  computed  AD).  Hence,  we  can  compute  B,  Fia.  18. 

rt,  and  DCB.  Thus,  the  parts  of  ACB  become  known  ....  When  the  given 
parts  are  two  angles  and  the  included  side,  find  the  third  angle  by  taking  the 
supplement  of  the  two  given.  Let  fall  a  perpendicular  from  one  extremity  of 
the  given  side  upon  the  opposite  side.  The  two  right  angled  triangles  thus 
formed  can  then  be  computed.  Thus,  if  A,  h,  and  C  are  given,  having  found  B, 
let  fall  CD.  The  triangle  ACD  has  the  angle  A  and  side  b  known,  whence  its 
parts  can  be  computed.  Having  computed  P  we  can  pass  to  the  triangle  CDB, 
^nd  knowing  P  and  B,  can  compute  it.   Thus  the  parts  of  ACB  be  '.ome  known. 


56  PLANE  TRIGONOMETRY. 

2nd.  W/ien  tico  of  the  given  parts  are  adjacent  and  one  separated;  i.  e  ,  when 
two  angles  and  a  side  opposite  one  are  given  ;  or  two  sides  and  an  angle  oppo- 
site one  are  given.  The  first  of  these  cases  is  viitually  the  same  as  the  last  given. 
To  solve  the  other,  let  fall  a  perpendicular  from  the  angle  between  the  given  sides, 
and  two  right  angled  triangles  will  be  formed  which  can  readily  be  computed. 
Thus  a,  b,  and  A  being  given,  and  CD  let  fall  from  C,  the  triangle  ACD  can 
first  be  computed,  and  then  CDB.  This  is  the  ambiguous  case,  but  it  is  easily 
determined.  Having  computed  P,  if  the  given  side  a  is  less  than  P  there  is  no 
solution ;  if  =  to  P,  on^  solution  (a  right  angled  triangle) ;  if  a  >  P  and  <  h,  there 
are  two  solutions,  i.  e.,  it  will  go  in  between  CD  and  AC,  and  also  beyond  CD ;  if 
a  >  P  and  also  >  h  there  is  only  one  solution,  as  it  will  not  go  in  between  CD 
and  AC. 

3rd.  When  the  three  given  parts  are  all  separated  from  each  oilier.  This  is  the 
case  in  which  the  three  sides  are  given  to  find  the  angles.  It  is  readily  solved 
by  letting  fall  a  perpendicular  from  the  angle  opposite  the  greatest  side,  upon 
that  side,  as  CD  upon  AB.  Then  compute  the  segments  AD  (which  call  m),  and 
DB  {n),  from  the  following  relation  (Part  II,  Ex.  12,  page  162) : 
m  +  71  (or  c)  :  h  +  a  :  :  b  —  a:m  —  n. 

Knowing  m  and  n,  the  angles  of  the  two  right  angled  triangles  ACD  and 
CDB  can  be  computed,  and  these  make  known  the  angles  of  ACB.    q.  e  d 

[Note.— A  few  additional  examples  are  here  given  which  the  pupil  can  use 
to  illustrate  the  theory  presented  in  (91).  If  more  are  needed  the  preceding 
can  be  used :  tliese  may  also  be  used  to  apply  the  methods  before  given.  Again, 
a  very  great  variety  and  number  of  examples  may  be  made  from  these  by  as- 
signing different  parts  as  known.] 


EXERCISES. 

1.  In  a  plane  triangle  BDF,  the  side  BF  is  123.75,  DF  500,  and  the 
included  angle  F  120°.    Find  the  angles  B  and  D,  and  the  side  BD. 

B  =  49°  12'  4";  D  =  10°  47'  56";  BD  =  572.006. 

2.  In  a  plane  triangle  ABC,  the  angle  A  is  70°  21',    the  angle  B 

54°  22',  and  the  side  BC  125.    Find  the  angle  C,  and  the  sides  AB 

and  AC. 

C  =  55°  17' ;  AB  =  109.1 ;  AC  =  107.88. 

3.  In  a  plane  triangle  ABC,  the  side  AB  is  98,  the  side  BC  95.12, 
and  the  angle  C  33°  21'.    Find  the  angles  A  and  B,  and  the  side  AC 

A  =  32°  14'  55";  B  =  114°  24'  5";  AC  =  162.33. 

4.  In  a  plane  triangle  DAC,  given  AD  =  450,  AC  =  309,  and  D  = 
27°  50',  to  find  the  other  parts. 

C  =  137°  9'  36",  or  42°  50'  24";  A  =  15°  0'24",  or  109°  19'  36"; 
OC  ^  171.36,  01  624.5. 


SOLUTION  OF  OBLIQUE  ANGLED  PLANE  TRIANGLES. 
COMPUTATION. 


It  Tvill  give  definiteness  to  the  stu- 
dent's thought,  if  he  first  sketch  the 
figure  geometrically.  Thus,  lay  off 
D  =  27°  50',  and  taking  AD  =  450, 
let  fall  the  perpendicular  AP. 

1st  To  compute  p. 
p=c  sin  D  =  450  sin 27' 50', 

log  450  =  2.653213 

log  sin  27°  50'  =  9.669225 

log  p  =  2.322438.    .-.  p  =  210.106, 


57 


Fig.  19. 


Knowing  p,  we  see  by  inspection  that  AC  can  lie  in  both  the  positions  AU 
'ind  AC,  and  hence  that  there  are  two  solutions. 

2nd.  To  compute  0^  from  the  triangle  ACP,  in  which  d  and  p  are  now 
snown. 

_P       210.106 
Sm  C  -  ^  -      3Qt^ 

log  210.106  =  2.322438 
log  309  =  2.489958 
log  sin  C     =  9.832480.     .-.  C'=  42°  50'  24",  and  C  =  137°  9'  33". 

3rd.  To  find  tlie  angle  A.  DAC  =  180°  -  (D  +  ACD)  =  180°  -  164°  59'  36"  = 
15°  0'  24".     DAC  =  180°  -  (D  +  ACD)  =  180°  -  70°  40'  24"  =  109°  19'  36". 

4th.  To  find  DC.  Compute  DP  and  CP  from  the  triangles  APD  and  APC. 
DP  -  CP  =  DC,  and  DP  +  CP  =  DC. 

5.  In  a  plane  triangle  ABC,  the  side  AB  is  460,  BC  is  340,  and  AC 
280.     Find  the  angles  A,  B,  and  C. 

A  =  47°  23'  16";  B  =  37°  18'  31";  C  =  95°  IS'  13". 

6.  The  sides  of  a  plane  triangle  are  40,  34,  and  25  feet  respect- 
ively ;  required  the  angles. 

38°  25'  ;20",  57°  41'  24",  83'  53'  16". 

7.  The  sides  of  a  plane  triangle  are  390,  350,  and  270  feet  respect- 
*vely;  required  the  angles. 

42°  22'  06",  60°  52'  33",  and  76°  45'  21". 

8.  Given  two  sides  of  a  plane  triangle  450  and  540,  and  the  in- 
cluded angle  80°,  to  find  the  remaining  parts. 

Angles,  56°  11',  43°  49';  and  the  side,  640.08. 

9.  Given  two  sides  of  a  plane  triangle  76  and  109,  and  the  in- 
cluded angle  101°  30',  to  find  the  remaining  parts. 

Angles,  30°  57'  30",  47°  32'  30";  and  the  side,  144.8. 


58 


PIAXE   TRIGONOMETRY. 


FOXTI0>S  OF  THE   A>GLES  OF  A  TRIA>GLE   IN   TERMS  OF  THE 

SIDES. 

9fl,  Tro]}. — A)uj  side  of  a  plane  triangle  equals  the  swn  of  the 
products  of  each  of  the  other  sides  into  the  cosine  of  the  angle  lohich  it 

makes  with  the  first  side, 

Dem.— In  the  first  figure  AB  =  AD  +  DB. 
But  AD  =  6  cos  A,  and  DB  =  a  cos  B.  /.  c  = 
&  cos  A  +  a  cos  B.  In  the  second  figure  AB  = 
AD  —  DB.  But  AD  =  6  cos  A,  and  DB  = 
a  cosCBD  =  a  (— cos  CBA)=  —  a  cos  B.  .-.  c 
=  b  cos  A  —  (—a  cos  B)  =  b  cos  A  +  a  cos  B. 
In  like  manner,  we  have  a  =  b  cos  C  +  c  cos- 
B,  and  b  =  a  cos  C  +  c  cos  A.  Collectmg  and 
ai-ranging. 

(1)  a  =b  cosC  +  c  cos  B ; 

(2)  b  =  acosC  +  c  cos  A ; 

(3)  c    =a  cos  B  +  b  cos  A.    q.  e.  d. 
Tie.  20. 


93,  Cor. — Tlie  square  of  any  side  of  a  plane  triangle  equals  the 
sum  of  the  squares  of  the  other  two,  minus  twice  their  rectangle  into 
the  cosiiie  of  their  included  angle. 

Dem. — From  (3)  {92),  we  have  by  transposmg  and  squaiing, 
a/*  cos'  B  =  c'  +  6"  cos'  A  —  2Z>c  cos  A  ;  and 
from  {85)  a"  sin'  B  =  6"  sm'  A. 

Adding,  a}  z=z  c^  +  6' —  2Z>c  cos  A. 

In  like  manner,  6'  =a^  -\-  c^  —  2ac  cos  B  ; 

and  c*  =  a'  +  6'  —  2aZ>  cos  C.    Q.  e.  d. 

94,  ScH. — These  formula  affor^l  another  means  for  finding  the  angles  of  a 
plane  triangle  when  the  sides  are  given.    Thus, 

(li)  cosA  = 

(2i)  cos  B  = 

(3.)cosC  = 

2ab 


J' 

+  c'- 

■a' 

2bc 

a' 

-t-c'- 

-&« 

2ac 

a' 

+  6»- 

c» 

These  foi-muloB  give  directly  the  natural  cosines  of  the  angles  in  terms  of 
the  sides.  To  adapt  them  to  logarithmic  computation,  we  transform  tliem  as 
follows : 

t>  +  c«  -  rt» 


Subtracting  each  member  of  (li)  from  unit}',  1  —  cos  A  =  1  — 

g«-y-c'4-  2hc_a''-{h-c)^_{a^{b-c)]  {a-{h-c)]_{a 
2be  ~        2bc         ~  2bc  ~ 


2bc 
b—c){a  +  e—b) 
26c~ 


SOLUTION  OP  OBLIQUE  ANGLED  PLANE  TKIANGLES.  59 

But  1  —  cos  A  =  2siiiHA  (57,  O) ;    and  Icltini;  j>  =  a  +  6  +  c,  ^(a  +  6  —  c) 
=  Ip _  Cj     and    i{a  +  C'-b)  =  ip  —  b.      Whence,      substituting,    2sin*iA  = 

2bc 


(U)  sin^A  =  i/^JL-L^^^ZLhl.    la  like  manner, 
(20siniB  =  |/ii^^^E«);and 

V  CIC 

In  a  manner  altogether  similar,  by  adding  each  member  of  (1')  t(i  unity,  and 
reducing,  we  get 

(Is)   cos  iA  =  a/^^  (^  -  ^)  ;  and  from  (2,)  and  (3.), 

(23)   cosiB  =  |/Si^HS; 
r  ac 

Dividing  (I2)  by  (I3),  (2^)  by  (2,),  and  (3^)  by  (3,),  we  have. 


^^^)  -^^=/^^^^ 


(3J    taniC  =  i/^p|ii^>. 
^       ii?(ii>-c) 


EXERCISES. 

[Note.— In  order  to  render  X\iQ%e,  formulcB  familiar,  and  to  give  the  student 
exercise  in  applying /<97'??i?i^,  a  few  examples  are  appended.  If  necessary,  any 
which  precede  can  be  used.] 

1.  The  sides  of  a  plane  triangle  being  40,  34,  and  25,  find  the 
angles. 

Solution. — By  natural  functions. 

Let  the  sides  be  represented  by  a,  6,  and  c  in  order,  and  the  angles  opposite 
by  A,  B,  and  C  ;  then 

Cos  A  =  '^^^r^-^  ^1136+623-1600 ^  _jog,,_    _.   '^  ^  33.  53.  ^^..^ 
2bc  1700 


60 


PLANE  TRIGONOMETRY. 


There  is  no  ambiguity  in  this  case,  since  the  cosine  is  +,  and  hence  the  angle 
is  <  90'. 

The  same  angle  is  found  by  logarithmic  computation,  thus  ; 


a  =  40  ...  log  =  1.602060 
6  =  34  ...  log  =  1.531479 
c  =  25    .  .  .log  =  1.397940 


p  =  99    ...  log  =  1.995635 

^  =  49.5  ...  log  =:  1.694605 

ip-a=    9.5  ...  log  =  0.977724 

^p  -b  =  15.5  .  .  .  log  =  1.190332 

^p-c  =  24.5  ...  log  =  1.389166 


log  {hP  -c)  =  1.389166 
\ogiip-  6)  =  1.190332 
a.  c.  log  b  =  8.468521 
a.  c.  log  c  =  8.602060 
2)1.650079 
T.825039 


log  sin  |A     =  9.825039. 
.-.  iA  =  41°  56'  38"  and  A  =  83°  53'  16". 

In  like  manner  the  other  angles  may  be  found. 

2.  The   sides  of  a  plane  triangle  being    6,  5,  and  4,  find   the 

angles. 

The  angles  are  82°  49'  09",  55°  46'  16",  and  41°  24'  35". 

3.  The  sides  of  a  plane  triangle  being  8601.5,  4082,  and  7068,  find 
the  angles. 

The  angles  are  54°  35'  12",  28°  4'  44",  and  97°  20'  4". 

4.  The  sides  of  a  plane  triangle  being  .51238G4,  .3538971,  and 
.3090507,  find  the  angles. 

The  angle  opposite  the  last  side  is  36°  18'  10".2. 


AREA  OF  PLANE  TRLiXGLES. 

Oo»  JPro2^' — The  area  of  a  plane  triangle  is  equal  to  half  the 
product  of  any  two  sides  into  the  sine  of  the  included  angle, 

Dem.— Let  ABC  be  any  plane  triangle,  and  b  and 
c  any  two  sides  with  A  as  the  included  angle.  From 
the  extremity  of  one  of  these  two  sides  remote  from 
A,  let  fall  a  perpendicular  p,  upon  the  othor  side 
Now, 

Area  ACB  =  ^pc. 


Fig.  21. 


But,   from   ACD,  ^  =  6  sin  A. 
ibc  sin  A-    q.  e.  d. 


Area  ACB 


06.  Cor. —  Tlic  area  of  a  plane  triangle  is  equal  to  the  square  root 
of  the  continued  product  of  half  its  periruter  into  half  its  perimeter 
minus  each  side  separately. 


AREA  OF  PLAME  TRIANGLES.  Gl 

Dem. — From  the  proposition,  and  since  sin  A  =  2sin  |A  cos  ^A,        we  have, 


^l^ea  =  ibc  sin  A  =  be  smiAcosi  A  =  be ^^^P^^tZ:^^  x  |/iME«I^ 


EXERCISES. 

1.  Given  two  sides  of  a  plane  triangle  125.81  and  57.05,  and  tlie 
included  angle  57°  25'.     Find  the  area. 

Area  =  3055.7. 

2.  Given  the  sides  of  a  plane  triangle  103.5  and  90,  and  the- 
included  angle  100°,  to  find  the  area. 

Area  =  45SG.74. 

3.  How  many  square  yards  are  there  in  a  triangle  whose  sides  are 
30,  40,  and  50  feet? 

AreazzQG^. 

4.  Find  the  area  of  a  triangle  whose  sides  are  20,  30,  and  40. 

Area  =  290.4737. 

5.  What  is  the  area  of  a  triangle  whose  sides  are  30  and  40,  and 
their  included  angle  28°  57'  ? 

Area  =  290.427. 

6.  What  is  the  number  of  square  yards  in  a  triangle,  of  which  the 
sides  are  25  feet  and  21.25  feet,  and  their  included  angle  45°  ? 

Area  -  20.8694. 

7.  Find  the  area  of  a  triangle  in  which  two  of  the  angles  are  80° 
and  60°  respectively,  and  the  included  side  32  feet. 

Area  =  679.33  square  feet. 

8.  Find  the  area  of  a  triangular  field  having  one  of  its  sides  45 
poles  in  length,  and  the  two  adjacent  angles,  respectively,  70°  and 
69°  40'. 

Area  =  1378.411  square  poles. 

9.  Find  the  area  of  a  triangular  piece  of  ground  having  two 
angles  respectively  73°  10'  and  90°  50',  and  the  side  opposite  the 
latter  75.3  poles. 

Area  =  748.03  square  poles. 


PRACTICAL  APPLICATIONS. 

[Note. — The  following  problems  are  inserted,  not  as  any  part  of  a  treatise 
npon  the  subject  of  trigonometry  as  pure  science,  but  as  affording  the  student 
rood  mental  exercise,  and  valuable  and  interesting  information.] 


62 


PLAKE  TKIGONOMETRY. 


1.  To  find  the  length  (in  miles)  of  a  degree  of  longitude  at  Ann 

Arbor,  Mich. 

Solution. — Let  NESQ  be  a  meridian  section  of 
the  earth,  EQ  the  equatorial  diameter,  and  EL  the  lati- 
tude of  Ann  Arbor,  42'  16'  48".3.  A  degree  of  longi 
tude  at  L  is  3^0  of  the  circumference  of  the  circle 
whose  radius  is  LD.  CL  the  radius  of  the  earth  at 
tliis  point  =  3957.*  Now  in  the  right  angled  ti-iangle 
LCD,  we  have  CLD  =  ECL  =  42°  16'  48".^3,  and  CL  = 
3957;  whence,  LD  =  CL  x  cos  42°  16'  4S".3,  and  LD 
=  2927.6.  .*.  A  degree  =  51.1  miles.  As  a  degree  in 
longitude  makes  4  minutes  di£Ference  in  time,  51.1 

miles  east  or  west  on  this  parallel  is  equivalent  to  4  minutes  difference  in  time. 

Query.— How  does  it  appear  from  the  above  solution  that  the  length  of  a 
degree  of  longitude  varies  as  the  cosine  of  the  latitude  ? 

2.  To  find  the  distance  of  a  planet  from  the  earth  at  any  par- 
ticular time. 


Solution — To  render  the  problem  as  simple  as  possible,  we  will  suppose  two 

observatories  on  the  same  meridian,  at  N,and 
N';  and  that  when  the  planet  P  is  on  the  same 
meridian,  the  angles  ZNP,  and  Z'N'P  (the 
zenith  distances)  are  measured.  With  these 
data  and  the  radius  of  the  earth,  CN,  CN', 
known,  the  problem  comes  quite  within  the 
scope  of  the  present  study.  The  process  is 
as  follows:  The  arc  NN' being  known,  the 
angle  NCN'  is  known.  Then  in  the  triangle 
NCN',  two  sides  and  tlie  included  angle  are 
known,  whence  the  other  parts  can  be  found. 
Now,  knowing  the  angles  PNC,  PN'C,  and 
CNN',  CN'N,  we  can  find  the  angles  PNN', 
PN'N.  This  affords  suflScient  parts  of  the  triangle  PNN'  to  determine  the  triangle, 
and  we  find  PN,  or  PN'.  Finally,  in  the  triangle  PNC,  we  know  PN,  NC,  and 
the  included  angles  whence  the  other  parts  can  be  computed.  But  PC  is  the 
distance  sought 


Fig.  23. 


3.  Suppose  in  case  of  the  moon,  the  angles  PNZ,  and  PN'Z',  being 
measured,  are  found  tp  be  respectively  44°  54'  21",  and  48°  42'  57", 
the  distance  between  the  points  of  observation  N  and  N'  is  92°  14', 
and  the  radius  of  the  earth  is  8956.2  miles;  find  the  distance  to  the 
moon. 

Distance  =  237,954.098  miles. 


•  The  equatorial  radinp  of  the  earth  is  3962.8  miles ;  but  in  consequence  of  the  flattening  in 
the  direction  of  the  polar  diameter  it  is  less  here. 

/ 


PRACTICAL   PEOBLEMS. 


63 


4.  Required  the  height  of  a 
hill  D  above  a  horizontal  plane 
AB,  the  distance  between  A  and 
B  being  equal  to  975  yards,  and 
the  angles  of  elevation  at  a  and 
B  being  respectively  15°  36'  and 
27°  29'. 


DC  =  587.61  yards 


5.  Find  the  area  of  a  regular  hexagon,  and  also  of  a  regular 
octagon,  whose  sides  are  each  10  feet. 

Areas,  259.8,  and  482.84  square  feet 

6.  Find  the  area  of  a  regular  pentagon,  and  also  of  a  regular  dec- 
agon, whose  sides  are  each  12  feet. 

Areas,  247.74,  and  1107.96  square  feet. 

7  Wishing  to  know  the  length  of  a  certain  pond  of  water,  1 
measured  a  line  100  yards  in  length,  and  at  each  of  its  extremities 
observed  the  angles  subtended  by  the  other  extremity  and  a  couple 
of  trees  at  the  extremities  of  the  pond.  These  angles  were,  at  one 
end  of  the  line,  32°  and  98°,  and  at  the  other,  37°  and  118°;  what 
was  the  length  of  the  pond  ? 


Draw  the  horizontal  line  AB  equal  to  100; 
make  the  angle  BAD  32%  BAC  98°,  ABC  37°, 
and  ABD  118°.  The  intersections  of  the  lines 
AC  and  BC,  AD  and  BD,  determine  the  extremi- 
ties of  the  pond;  the  straight  line  CD  is  the 
length  of  the  pond. 

CD  =  161.868  yards. 


8.  The  distances  AB,  AC,  and  BC,  between 
the  points  A,  B,  and  c,  are  known  ;  viz.,  AB  = 
800  yds.,  AC  =  600  yds.,  and  BC  =  400  yds. 
From  a  fourth  point  P,  the  angles  APC  and  BPC 
are  measured;  viz.,  APC  =  33°  45', 
and  BPC  =  22°  30'. 

■Requir'id  the  distances  AP,  BP,  and  CP. 

r  AP  =    710.193  yds. 

Distances,^  BP  =    934.291  yds. 

t  CP  =  1042.522  yds. 


64  PLANE  TRIGONOMETRY. 

Sug's. — Conceive  tlie  circumference  passed  through  A,  B,  and  P,  and  AD  and 
DB  drawn.  In  the  triangle  ADB,  angle  DAB  =  the  given  angle  DPB,  and  DBA 
—  APD.  Ilence,  all  the  parts  of  triangle  ADB  can  be  found.  Again,  since  the 
sides  of  the  tiiangle  ACB  are  given,  its  angles  can  be  found.  Then,  since  angle 
CAB  —  DAB  =  CAD,  there  are  two  sides  and  the  included  angle  known  in 
triangle  ACD  ;  whence  angle  ACD  can  be  found.  Thus  we  reach  the  ti'iangle 
ACP ,  in  which  there  are  now  known  AC  and  the  angles. 

9..  From  the  top  of  a  mountain,  three  miles  high,  the  angle  of 
depression  of  a  line  tangent  to  the  earth's  surface  is  taken,  and 
found  to  be  2°  13'  27".  What  is  the  diameter  of  the  earth,  considered 

as  a  sphere  ? 

Arts.  7946.28  miles. 

10.  Taking  the  sun's  mean  apparent  diameter  as  32'  3".4,  and  his 
distance  from  the  earth  91,430,000  miles,  show  that,  if  his  centre 
were  coincident  with  the  earth's,  his  body  would  extend  in  all  direc- 
tions nearly  200,000  miles  beyond  the  moon.     (See  Ex.  3.) 

Sun's  diameter  =  852,574  miles. 

11.  Assuming  the  height  of  the  Great  Pyramid  to  be  486  feet,  how 
far  off  may  it  be  seen  across  the  desert  ? 

Ans.,  27  miles. 

12.  "WTiat  was  the  perpendicular  height  of  a  balloon,  when  its 
angles  of  elevation  were  35°  and  64°,  as  taken  by  two  observers  on 
the  same  level,  at  the  same  time,  both  on  the  same  side  of  it,  and 
in  the  same  vertical  plane ;  the  distance  between  the  two  observers 

being  880  yards  ? 

Ans.y  935.757  yards. 

13.  Given  two  sides  of  a  parallelogram  60  and  80,  and  a  diagonal 
100.  Is  this  the  longer  or  shorter  diagonal  ?  What  is  the  other  ? 
What  are  the  angles  of  the  parallelogram  ? 

14.  A  balloon  being  directly  over  one  of  two  towns  standing  on 
uhe  same  horizontal  plane,  at  a  distance  of  eight  miles  from  each 
ether,  the  angle  of  depression  to  the  more  remote  town  was  observed 
by  the  aeronaut  to  be  10°.    What  was  the  height  of  the  balloon  ? 

Ans.,  1.41  miles. 

15.  The  most  recent  observations  make  the  sun's  horizontal  par- 
allax 8 ''.94,  and  the  earth's  equatorial  radius  3962.8  miles.  Show 
that  the  distance  of  the  sun  from  the  earth  is  nearly  as  given  in  Ex. 
10,  instead  of  95  millions  of  miles,  as  it  has  been  heretofore  con- 
sidered. 


CHAPTER  n. 

SrHEBICAL    TBIGONOMETBY^ 

INTR  OD  UCTION. 

PROJECTION  OF  SPHERICAL  TRIANGLES. 

97.  To  Project  a  Spherical  Triangle  on  a  plane  surface 
is  to  draw  the  triangle  on  that  surface  so  that  it  will  present  the 
same  appearance  to  the  eye,  situated  at  a  particular  point,  as  when 
drawn  on  the  surface  of  a  sphere. 

98.  The  Simplest  Method  of  projecting  a  spherical  triangle 
is  to  project  it  on  the  j^lane  of  one  of  its  sides,  the  eye  being  supposed 
situated  in  the  axis  of  the  sphere  perpendicular  to  this  plane,  and  at 
an  infinite  distance  from  it.  The  plane  is  called  the  Plane  of  Pro- 
jection ;  and  its  intersection  with  the  sphere  is  called  the  Primitive 
Circle,  and  is  the  base  of  the  hemisphere  on  which  the  triangle  is 
conceived  as  situated. 


99.  Fundamental  Propositions. — 1st.  Wientlie  ixirts  of 
a  spherical  triangle  are  each  conceived  as  less  than  180°,  any  such 
triangle  can  he  represented  on  a  hemisphere, 

2d.  The  primitive  Circle  has  its  axis,  and  consequently  its  pole, 
projected  at  its  centre. 

3d.  The  semi-circumference  of  any  circle  of  the  sphere,  perpendic- 
ular to  the  Primitive  Circle,  is  p^rojected  in  the  chord  representing 
the  intersection  of  the  circles  ;  and,  if  the  perpendicular  circle  he  a 
great  circle,  its  semi-circumference  is  projected  in  a  diameter  of  the 
Primitive  Circle. 


6G 


SrHERICAL   TRIGONOMETRY. 


Ill's. — These  propositions  are  direct  cousequences  of  the  fundamental  con- 
ception.^  Thus,  let  ABA'B'  represent  the  base  of  the  hemisphere  on  which  the 

triangle  is  conceived  as  situated.  This  is  the 
Primitive  Circle,  and  the  e3'^e  is  supposed  situated 
at  an  infinite  distance,  and  in  a  line  perpendicular 
to  the  plane  of  the  paper  at  P.  The  pole  of  the 
Primitive  Circle  being  in  this  line  is  projected 
(seen  as)  at  P.  As  all  great  circles  perpendicular 
to  the  Primitive  Circle  pass  through  its  pole  and 
include  its  axis,  the  eye  is  in  all  such  planes,  and 
any  lines  of  these  planes,  as  the  semi-circumfer- 
ences of  the  great  circles  in  which  they  intersect 
the  sphere,  are  projected  (appear  to  the  eye)  as 
diameters  of  the  Primitive  Circle.  Moreover, 
since  the  eye  is  at  injinily,  it  is  to  be  conceived  as  in 
the  plane  of  any  small  circle  which  is  perpendicular  to  the  primitive,  and  which 
is  therefore  projected  in  a  chord,  as  CC. 


PROJECTION  OF  RIGHT  ANGLED  SPHERICAL  TRIANGLES. 


100.  JProb,  1. — To  project  a  right  angled  spherical  triangle  on 
the  plane  of  one  of  its  sides,  luhen  tlic  tiuo  sides  about  the  right  angle 
are  given.* 

Solution. — Let  the  angles  of  the  tnangle  be  represented  hy  A,  B,  and  C,  A 
Let  the  sides  opposite  these  angles  respectively  be  rep- 
resented by  a,  h,  and  c ;  whence  b  and  c  are  the 
given  sides.  Draw  the  primitive  circle  and  the 
diameters  BB',  NN'  at  right  angles  to  each  other. 
From  B  lay  off  BA  =  c.\  Let  the  right  angle  be 
at  A;  whence  the  side  b  is •  perpendicular  to  the 
primitive  circle,  and  projected  in  the  diameter  A  A'. 
To  project  the  vertex  C,  conceive  the  semi-circum- 
ference, of  which  A  A'  is  the  projection,  to  revolve 
on  AA'  until  it  falls  upon  the  semi-circumference 
A'BA,  then  will  the  point  C  f\\ll  at  d.  Hence  make 
kd  =  b.  In  like  manner  revolving  the  semi-cir- 
cumference, of  which  AA'  is  the  projection,  until 
it  falls  upon  A'B'A,  the  point  C  will  fall  at  d'. 
Hence  make  M'  =  b.  The  point  0  will  de- 
scribe the  semi-circumference  of  a  small  circle 
perpendicular  to  the  primitive  circle,  and  whose  projection  is  dd'.     Now,  as  the 


Fig.  25. 


*  In  this  treatise  the  discussions  embrace  only  such  triangles  as  have  each  part  less  than 
180°. 

t  For  the  purposes  for  which  we  shall  use  these  projections,  an  arc  can  be  laid  olT  with 
sufficient  accuracy  by  observing  its  relation  to  90°,  60°,  30°,  or  some  aliquot  part  of  the  circum- 
ference, which  is  readily  obtained  on  geometrical  principles. 


PROJECTION   OF  EIGHT  ANGLED   SPHEEICAL  TRIANGLES. 


67 


projection  of  the  vertex  of  the  triangle  C  is  at  the  same  time  in  AA'  and  cld\  it 
must  be  at  their  intersection.  Finally,  the  hypotenuse  a  is  projected  in  a  curve 
passing  through  B,  C,  and  B',  since  two  great  circles  intersect  at  the  extremities 
of  a  diameter.  This  curve  is  really  an  ellipse,  but  for  our  present  purpose  it 
may  be  considered  as  the  arc  of  a  circle  passing  through  B,  C,  and  B'.  Tiiere- 
fore,  BAG  is  the  projection  required. 

Queries. — Will  a  solution  of  this  problem  be  possible  for  all  values  of  J  and 
c  ?    How  does  it  appear  from  the  projection  ? 


EXAMPLES. 

1.  Having  given  h  =  110°,  and  c  =  60°,  to 
project  the  triangle.     See  Fig.  26. 

2.  Having  given  h  =  50°,  and  c  =  130°,  to 
project  the  triangle. 

3.  Having  given  I?  =  90°,  and  c  =  30°,  to    \ 
project  the  triangle. 

4.  Haying  given   b  =  90°,  and  c  =  90°,  to 
project  the  triangle. 


101.  JProb.  2, — To  ^^J'oject  a  right  angled  splicricaJ  iriangh 
tolien  the  liypotemise  and  one  side  ar'e  given. 


Solution. — Using  the  common  notation,  let  c  represent  the  known  side. 
Drawing  the  primitive  circle  and  the  conjugate*  diameters  BB',  NN',  layoff 
BA  =  c,  and  draw  the  diameter  AA'.  .  The  projec- 
tion of  b  will  lie  in  AA',  and  the  projection  of  the 
vertex  C  will  fall  somewhere  in  this  line.  Now 
the  arc  a  lies  in  a  semi-circumference  passing 
through  B  and  B'.  Conceive  this  semi-circumfer- 
ence to  revolve  on  BB',  as  an  axis,  till  it  coincides, 
first,  with.BNB',  and  then  with  BN'Bl  The  point 
G  will  trace  the  semi-circumference  of  a  small 
circle  perpendicular  to  the  primitive  circle,  and 
whose  projection  is  dd'.  Hence,  making  Bd  =  Bd' 
=  a,  and  drawing  dd\  the  projection  of  the  vertex 
C  lies  at  the  same  time  in  AA'  and  dd\  and  is  there- 
fore at  their  intersection.  Passing  an  arc  of  a 
circle  (strictlj^  an  ellipse)  through  BCB',  we  have 
ABC,  the  projection  desired. 


Fig.  27. 


*  Two  diakieters  of  a  circle  which  are  at  right  angles  to  each  other  arc  called  ConjugaU 
DiameUrs. 


68 


SPHERICAL  TRIGONOMETRY. 


Queries. — Will  a  solution  of  this  problem  be  possible  for  all  values  of  a  aud 
e  ?  If  a  had  been  greater  than  c  in  the  above  case,  would  the  solution  have 
been  possible  ?  Will  dd'  and  AA'  always  intersect,  whatever  may  be  the  relative 
values  of  a  and  c  ? 

EXAMPLES. 

1.  Having  given  c  =  75°,  and  a  =  64°,  to  project  the  triangle. 

2.  Having  given  c  =  45°,  and  a  =  136°,  to 
project  the  triangle. 

3.  Having  given  b  =  110°,  and  a  =  85°,  to 
project  the  triangle.     See  Fig.  28. 

4.  Having  given  b  =  110°,  and  a  =  120°, 
to  project  the  triangle. 

■   5.  Having  given  c  =  90°,  and  a  =  75°,  or 
a  =  120®,  to  project  the  triangle. 


Fig.  28. 


Query. — If  one  side  of  a  right  angled  spherical 
tiiangle  is    90°,  what  must    the   hypotenuse  be  ? 


Why? 


102,  JProb,  3. — To  2^J^oject  a  right  angled  spherical  triangle 
when  an  ohlique  angle  and  the  hypotenuse,  or  the  oblique  angle  and 
the  adjacent  side  are  given. 

Solution. — Draw  the  primitive  circle  and  the  conjugate  diameters  BB',  NN' 
as  usual.     To  construct  the  given  angle   B,  we  observe  that  this  angle  is 

measured  by  an  arc  of  the  great  circle  which  is 
projected  in  NN'.  Hence,  lay  oflf  Hd  =  N^'  =  B , 
and  draw  rfcf  ;  then  is  NO  the  projection  of  the 
arc  which  measm'es  B,  and  the  projection  of 
a  lies  in  tlie  arc  passing  through  BOB'.*  Having 
the  angle  B  projected,  if  the  hypotenuse  a  is  the 
other  given  part,  find  the  projection  of  C  by 
taking  Be  =  Be'  =  a,  and  drawing  ee'.  Com- 
plete the  projection  by  drawing  AC  through  P. 
When  the  adjacent  side  c  is  given,  take  BA  =  c, 
and  draw  AC  as  before.  [The  student  will  be 
able  to  give  the  reasons.] 

103,  ScH.— As  OP  is  the  cosine  of  the  angle 
ABC,  the  point  0  may  be  found  by  measuring 


*  A?  has  been  remarked,  this  arc  is  really  a  semi-ellipse.    This  fact,  together  with  the 
method  of  constructing  the  semi-ellipse,  and  thus  getting  the  correct  projection  of  the  hypot 


PROJECTION   OF  EIGHT  ANGLED   SPHERICAL  TRIANGLES. 


69 


from  P  (towards  N  if  B  <  90%  and  towards  N'  if  B  >  90°)  a  distance  equal  to 
the  natural  cosine  of  B. 


Query. — Is  the  solution  of  this  problem  possible  for  all  values  of  the  hj'pot- 
enuse  or  adjacent  side,  and  the  angle? 

EXAMPLES. 

1.  Having  given  B  =  65°,  and  the  hypotenuse  a  =  120°,  to  proiect 
the  triangle.     See  Fiff.  30. 

2.  Having  given  C  =  45°,  and  the  adjacent 
side  b  =  50°,  to  project  the  triangle. 

Sug's. — Project  the  angle  C  as  the  angle  B  of 
the  preceding,  and  lay  off  5  =  50°  from  its  vertex 
on  the  circumference  of  the  primitive  circle. 

3.  Having  given  C  =  170°,  and  hypote- 
nuse a  =  160°,  to  project  the  triangle. 

4.  Having  given  B  =  150°,  and  c  =  40°,  to  project  the  triangle. 


104,  JProb.  4, — To  2^^oject  a  right  angled  spherical  triangle 
when  an  ohlique  angle  and  side  opposite  are  giveii. 


SoLUTiox. — Project  the  given  angle  at  B,  Figs.  31,  32,  as  in  the  last  problem. 
Then,  from  an}'  point  in  the  circumference  of  the 
primitive  circle,  as  N,  take  NO',  in  the  diameter 
passing  through  that  point,  equal  to  the  projection 
of  the  given  side.  (This  is  done  by  taking  Hd  = 
Nd'  =  b,  as  in  Prob.  1,  and  drawing  del').  Now, 
with  P  as  a  centre,  and  radius  PO',  describe  a  cir- 
cumference cutting  BOB'.  One  extremity  of  the 
"given  side  h  will  be  projected  in  this  circumference, 
since  this  circumference  contains  the  projections  of 
all  the  points  in  the  surface  of  the  hemisphere 
which  are  at  a  distance  h  from  the  circumference 
BNB'N'.    But  the  vertex  C  is  also  projected  in  the 


cnuse,  belong  to  a  treatise  on  Conic  Sections.  In  this  case,  BB'  and  20  P  are  the  axes  of  the 
ellipse,  and  the  curve  can  be  constructed  by  takin<?  BP  as  a  radius,  and  strikinor  arcs  from  0 
as  a  centre,  cutting  BB'-  These  intersections  are  the  foci  of  the  required  ellipse.  Then  take 
a  string  equal  in  length  to  BB',  and,  fastening  its  ends  to  the  foci,  place  a  pencil  against  the 
string,  and  keeping  the  string  tight,  carry  the  pencil  around  the  curve. 


70 


SPHERICAL  TEIGONOMETRY. 


Fig.  32. 


ai'c  BOB' ;  hence,  it  must  be  at  the  intersections 
C,  C  Drawing  the  projections  of  the  perpendic- 
ulars CA,  and  C'A',  the  projection  is  complete. 

Queries. — When  will  there  be  two  triangles 
fulfilling  the  given  conditions  ?  "When  but  one  ? 
When  none  ?  When  there  is  but  one  triangle  what 
kind  of  a  triangle  is  it?  If  B  >  90%  must  b  be 
greater  or  less  tlian  B  in  order  to  have  two  solu- 
tions ?  If  B  <  90%  how  is  it  ?  If  B  >  90°,  can  b  be 
less  ?    If  B  <  90%  can  o  >  90="  ? 


EXAMPLES. 

1.  Given  C  =  120°,  c  =  150,  to  project  the  triangle.     See  Fig.  33. 

2.  Given  c  =  80°,  c  =  60°,  to  project  the  triangle.     See  Fiff.  3-4. 


Fig.  33. 


3.  Given  B  =  70^ 

4.  Given  B  =  6-4°, 


Fig.  34. 


T0°,  to  project  the  triangle.     See  Fig.  35. 
75°,  to  project  the  triangle.     See  Fig.  36. 


A  \ 

Fig.  35.  Fig.  36. 

5.  Given  b  =  160°,  b  =  110°,  to  project  the  triangle. 


PROJECTION   OF   OBLIQUE   ANGLED    SPHERICAL  TRIANGLES. 


71 


105,  Sen. — When  the  given  parts  are  the  two  oblique  angles,  the  projection 
is  most  readily  effected  by  first  computing  one  of  the  sides.  The  projection  in 
this  case  will  be  considered  in  connection  with  the  numerical  solution  of  the 
case,  in  the  next  section. 


PROJECTION  OF  OBLIQUE  ANGLED  SPHERICAL  TRIANGLES. 
lOG.  Proh,  1, — To  project  a  sjylierical  trimigle  when  two  sides 


Project  the  given  angle 


and  the  included  angle  are  given. 

Solution. — Let  a,  e,  and  B  denote  the  given  parts. 
B  at  some  point  in  the  circumference  of  the  primi- 
tive circle,  as  B.  Lay  off  one  of  the  given  sides,  as 
c,  from  B  on  BNB'.  Let  BA  =  c.  Determine  the 
extremity  C  of  the  projection  of  the  other  given  side 
a,  as  in  Prob.  2,  etc.,  and  drawing  the  diameter 
AA',  pass  tlie  arc  through  ACA' ;  BCA  is  the  pro- 
jection sought. 

Query. — Is  this  projection  always  possible,  what- 
ever the  relative  magnitude  of  the  given  parts  ? 


EXAMPLES. 

1.  Given  A  =  130°,  c  =  85°,  and  b  =  100°,  to  project  the  triangle. 

2.  Given  c  =  40°,  a  =  37°,  and  I?  =  80°,  to  project  the  triangle. 


107.  JProb.  2, — To  project  a  spherical  triangle  tv7ie7i  tivo  sides 
and  an  angle  opposite  one  of  tliem  are  given. 

Solution. — Let  the  given  parts  be  a,  5,  and  B.  Make  the  projection  upon 
the  plane  of  tlie  iinknowii  side  c.  Thus,  drawing  the  primitive  circle  and  the 
conjugate  diameters  BB',  NN',  conceive  c  as  projected  from  B  on  the  arc  BNB', 
and  project  the  given  angle  B  as  in  preceding 
problems.  On  the  arc  BB',  take  BC  =  the  pro- 
"jection  of  the  given  adjacent  side  a.  To  deter- 
mine the  projection  of  the  opposite  side  5,  describe 
a  circle  about  P,  as  a  centre,  with  a  radius  PC. 
Through  C  draw  PD,  and  taking  Dd  =  Dd'  =  b 
draw  dd\  Tlirough  the  intersections  o,  o'  of  dd! 
with  the  circumference  of  the  small  circle,  draw 
the  radii  PA,  PA'.  Finally,  passing  arcs  througli 
the  points  A,  C,  A",  and  A',  C,  A'",  BAC,  and 
BA'C  are  the  projections  of  triangles  which  fulfill 
the  given  conditions.    The  projections  of  B  an(\ 


72 


SPHERICAL  TRIGONOMETRY. 


a  were  made  upon  principles  pre\iously  established ;  and  it  only  remains  to 
show  that  AC,  and  A'C  are  projections  of  b.  Since  by  construction  DL  is  the 
projection  of  an  arc  equal  to  b,  the  projections  of  arcs  of  great  circles  connect- 
ing D  with  0  and  o'  are  projections  of  arc^  equal  to  b.  But  the  figure  OoP 
—  ACP,  and  Do'P  =  A'CP ;  therefore  AC  =  A'C  =  Do  =  the  projection  of  b. 


108.  ScH.— It  is  evident  that  this  problem  has  one  solution,  two  solutions,  or 
no  solution,  according  to  the  value  of  6  as  compared  with  a  and  B.  Thus,  if  the 
projection  of  6  =  DC,  o  and  o'  coincide,  there  is  but  one  solution,  and  the  tri- 
angle is  right  angled  at  A  (which  in  this  case  falls  at  D).  Also,  if  the  projection 
of  b  is  intermediate  in  value  between  DC  andon/^  one  of  the  arcs  BC,  B'C,  there 
is  onXjone  solution.  If,  however,  as  in  the  figure  given,  the  projection  of  b  is 
intermediate  in  value  between  DC  and  both  BC  and  B'C,  there  are  tico  solutions. 
Finally,  if  6  is  given  of  such  value  that  the  chord  M  does  not  touch  the  small 
circle,  there  is  no  solution.  The  latter  case  occurs  when  B  <  90%  if  the  projec- 
tion of  Z>  <  DC  ;  and  when  B  >  90%  if  the  projection  of  J  >  DC,  as  will  appear 
from  Figa.  38,  39. 

We  may  observe,  also,  that  there  are  tico  solutions  when  o  and  o'  both  fall  on 
the  same  side  of  BB  as  c  ;  one  solution  when  o  and  o'  coincide,  and  when  they 
fall  on  opposite  sides  of  BB'  ;  and  no  solution  when  o  and  o'  are  imaginary,  i.  c, 
when  dd'  does  not  touch  the  small  circle,  or  when  both  fall  on  the  opposite  side 
of  BB'  from  c. 


EXAMPLES. 


Fis.  39. 


1.  Given  B  =  110°,  a  =  120°,  and  5  =  83°, 
to  project  the  triangle.     See  Fig.  39. 

2.  Given  B  =  110°,  a  =  120°,  and  ^>  =  130°, 
to  project  the  triangle. 

3.  Given  C  =  64°,  a  =  120°,  and  c  =  75°, 
to  project  the  triangle. 

4.  Given   C  =  80°,  b  =  60°,  and  c  =  40°, 
to  project  the  triangle. 

5.  Given  0  =  112°,  b  =  75°,  and  c  =  150°, 
to  project  the  triangle. 


109.  I^rob.  3.— To  project  a  spherical  triangle  when  the  three 
sides  arc  given. 


SOLUTION   OF  BIGHT  ANGLED    SPHERICAL  TRIANGLES. 


73 


Solution.— Drawing  the  primitive  circle  and  the  conjugate  diameters,  as 
nsual,  take  BA  =  c,  the  side  on  the  plane  of  which  it  is  proposed  to  project  the 
triangle.  Take  Be  =  Be'  =  a  and  draw  ee' ;  then 
as  before  shown,  the  projection  of  the  vertex  C 
lies  in  ee'.  In  like  manner  taking  ^d  —  Ad'  =  b, 
and  drawing  dd'  the  projection  of  C  lies  in  dd[. 
The  intersection  of  the  chords  ee'  and  dd'  is  there- 
fore the  projection  of  the  vertex  C.  Finally, 
passing  arcs  through  AC  A'  and  BCB',  the  projec- 
tion is  complete. 

Queries.— How  does  the  projection  show  the 
impossibility  when  the  sum  of  the  three  sides  is 
greater  than  3G0°  ?  How  when  the  sum  of  two 
sides  is  less  than  the  third  side  ?  Fiu.  40. 


EXAMPLES. 

1.  Given  a  =  100°,  b  =  80°,  and  c  =  68°, 
to  project  the  triangle. 

2.  Given  a  =  108°,  b  =  120°,  and  c  =  25°, 
to  project  the  triangle.     See  Fig.  41. 

3.  Given  a  =  120°,  I?  =  65°,  and  c  =  40°, 
to  project  the  triangle. 

4.  Given  a  =  150°,   b  =  140°,  and  c  = 
170°,  to  project  the  triangle. 


Fio.  41. 


SECTION  I. 


SOLUTION  OF  RIGHT  ANGLED  SPHERICAL  TRIANGLES. 

110.  Splierical  Trigononietry  treats  of  the  relations  be- 
tween the  trigonometrical  functions  of  the  sides  and  angles  of 
spherical  triangles,  and  of  the  solution  of  such  triangles  by  means  of 
these  relations. 

111.  ScH.— In  the  present  treatise  we  shall  confine  our  attention  to  triangles 
none  of  whose  sides  or  angles  exceed  180°. 

112.  The  Sioc  Parts  of  a  spherical  triangle  are  the  three 
sides  and  the  three  angles :  any  three  of  these  being  given,  the  others 
may  be  found. 


74  SPHERICAL  TRIGONOMETRY. 

113,  SCH.— The  last  statement  involves  the  assertion  that  the  three  angles 
of  a  spherical  triangle  determine  the  sides,  whereas  we  are  accustomed  to  say 
in  Plane  Trigonometry,  that,  at  least  one  given  part  of  a  plane  triangle  must  be 
a  side,  in  order  that  the  triangle  may  he  determined.  Tliere  is  really  no  such 
difference  as  these  two  statements  imply.  For  example,  if  we  have  the  angles 
of  a  plane  triangle  given,  we  know  the  ratios  of  the  sides  to  each  other,  since 
the  sides  are  to  each  other  as  the  sines  of  the  angles  opposite  ;  but  we  cannot 
determine  the  absolute  mines  of  the  sides.  This  is  in  accordance  with  the  state- 
ment that  mutually  equiangular  plane  triangles  are  similar  figures  (not  neces- 
sarily equal).  Now  these  are  exactly  the  facts  in  the  case  of  spherical  triangles, 
if  ue  do  not  limit  them  to  the  same  or  equal  spheres.  Thus,  the  angles  of  a  spheri- 
cal triangle  being  given  as  4S'  30',  125°  20',  and  62°  54',  we  solve  tlie  triangle  by 
the  rules  of  spherical  trigonometry  and  find  that  the  sides  are  56^  39'  30", 
114°  29'  58",  and  83°  12'  6".  But,  so  long  as  the  radius  of  the  sphere  is  unknown, 
these  results  are  merely  ikQj:£ltUi£^  values  of  the  sides,  not  their  absolute  lengths. 
Moreover,  consider  two  concentric  spheres  whose  radii  are  m  and  n.  Now,  any 
triangle  being  constructed  on  the  one,  if  planes  are  passed  through  its  sides 
intei*secting  at  the  common  centre,  their  intersection  with  the  surface  of  the 
other  sphere  will  form  a  triangle  mutually  equiangular  with  the  first,  and  any 
one  side  of  the  one  triangle  is  to  the  corresponding  side  of  the  other,  as  the  radii 
of  the  spheres  ;  hence  the  homologous  sides  are  proportional.  We  see,  therefore, 
that  to  determine  absolutelj^  a  spherical  triangle,  it  is  necessary  to  know  one  of 
the  sides  in  linear  extent  as  well  as  angular  measure,  or,  what  is  equivalent, 
the  radius  of  the  sphere  must  be  known. 

114,  TJie  Species  of  a  part  of  a  spherical  triangle  is  deter- 
mined by  its  relation  to  90°.  Two  parts,  both  greater  or  both  less 
than  90°,  are  of  the  same  S2)erAes  ;  two  parts,  one  of  which  is  greater 
and  the  other  less  than  90°,  are  of  different  sjjecies.  Thns,  two  parts 
which  are  58°,  and  63°,  respectively,  are  of  the  same  species ;  two 
which  are  160°,  and  115°,  are  of  the  same  species ;  two  which  are 
120°,  and  48°,  are  of  different  species. 

1  IS,  Napier^ s  Five  Ci7*ciilar  JParts  are  ^jwo  sides  of  a 
rij/ht  anfjled  spherical  triangle  about  the  right  an^le,  and  the  comjile- 
ments  of  the  h}i3otenuse  and  of  the  oblique  angles.  These  terms  are 
merely  conventional,  and  are  applied  exclusively  to  right  angled 
triangles. 

III.— In  a  spherical  triangle  ABC,  right  angled  at  A,  the 
sides  h  and  c,  the  complement  of  hypotenuse  «,  and  the  com- 
plements of  the  angles  B  and  C,  are  the  circular  parts.  We 
may  designate  them  5,  c,  comp  a,  comp  B,  and  comp  C. 

[It  is  essential  that  this  nomenclature  and  the  statements 
of  the  two  following  paragraphs  be  clearly  understood,  and 
firmly  fixed  in  the  mind,  in  order  that  the  phraseology  of 
^iG  42.  ^^^  fundamental  rules  may  be  intelligible.] 


SOLUTION   OF  EIGHT   ANGLED   SPHERICAL  TRIANGLES.  75 

116*  When  five  things  occur  in  succession,  as  it  were  in  a  circle, 
like  the  circular  parts  b,  c,  comp  B,  comp  a,  and  comp  c,  in  the 
preceding  figure  (no  account  being  made  of  A),  it  will  be  observed, 
that,  taking  any  three  at  pleasure,  one  of  the  three  may  always  be 
selected  which  lies  adjacent  to  each  of  the  others,  or  separated  from 
each  of  them.  Of  the  three  parts  thus  considered,  the  3Iiddle  JPart 
is  the  one  which  has  the  other  two  adjacent  to  or  separated  from  it; 
while  the  latter  are  called  the  Uxtrenies,  adjacent  or  02)2^osite,  as 
the  case  may  be. 

Ill's. — Let  the  three  parts  under  consideration  be  comp  a,  b,  and  c;  comp  a 
is  tlie  middle  part,  and  b  and  c  are  the  o2yposite  extremes.  If  b,  c,  and  comp  C 
are  under  consideration,  b  is  the  middle  part,  and  c  and  comp  C  are  the  adjacent 
extremes. 

117»  Sen. — In  solving  a  right  angled  spherical  triangle,  there  are  always 
three  parts  under  consideration  at  once,  viz.,  the  two  given  parts  and  the  part 
sought,  no  mention  being  made  of  the  right  angle.  Now,  tlie  first  question  to  be 
settled  in  oi-der  to  a  solution  is,  WJdcJi  of  the  three  parts  under  consideration  is  the 
middle  part^  and  are  tlie  extremes  opposite  or  adjacent?  Beginners  are  very  liable 
to  make  mistakes  by  failing  to  use  the  complements  of  the  proper  parts ;  or  by 
not  correctly  distinguishing  the  middle  part,  and  the  character  of  the  extremes, 
as  opposite  or  adjacent.  The  student  should  practise  upon  such  simple  exer- 
cises as  the  following  until  the  questions  can  be  answered  instantly^  and  with- 
out mistake. 


EXERCISES. 


1.  In  Fig.  42,  given  a  and  c,  to  find  C.  What  are  the  circular  parts 
under  consideration  ?  Which  is  the  middle  part  ?  Are  the  extremes 
adjacent  or  opposite  ? 

Answers. — The  circular  parts  are  comp  a,  c,  and  comp  C.  c  is  the 
middle  part,  and  the  extremes  are  opposite. 

2.  Having  C,  and  a  given,  to  find  h.  What  are  the  circular  parts  ? 
Which  is  the  middle  part  ?    Are  the  extremes  adjacent  or  opposite  ? 

3.  Having  c,  and  h  given,  to  find  B.  What  are  the  circular  parts  ? 
Which  is  the  middle  part  ?    Are  the  extremes  adjacent  or  opposite  ? 

4.  Having  a,  and  1)  given,  to  find  C.  What  are  the  circular  parts  ? 
Which  is  the  middle  part  ?    Are  the  extremes  adjacent  or  opposite? 

0.  Having  B  and  C  given,  to  find  I.  What  are  the  circular  parts  ? 
Which  is  the  middle  part  ?    Are  the  extremes  adjacent  or  opposite  ? 


76 


SPHERICAL  TRIGONOMETRY. 


6.  What  are  the  opposite  extremes  when  h  is  the  middle  part? 
What  the  adjacent  extremes  ?  Which  are  the  opposite  and  which  the 
adjacent  extremes  when  c  is  ^the  middle  part"^  When  comp  B  is  the 
middle  part  ?  When  comp  c  is  the  middle  part  ?  When  comp  a  is 
the  middle  part  ? 

7.  What  part  is  middle  part  to  comp  C  and  c  as  adjacent  ex- 
tremes?   As  opposite  extremes? 

Ans.,  l,  and  none, 

8.  In  Fig.  43,  M  being  the  right  angle, 
what  are  the  circular  parts  ?  Given  O  and  m, 
to  find  0.  What  are  the  circular  parts  under 
consideration  ?  Are  the  extremes  adjacent  or 
opposite  ? 

9.  What  are  the  opposite  extremes  to  comp 
O  ?  What  the  adjacent  ?   To  comp  m  ?  To  o? 

Pig.  43.  Ton? 


NAPIER'S  RULES. 

118,  Hule  !•  ^rop* — In  any  right  angled  sjyJierical  triangle^ 
the  sine  of  the  middle  part  equals  the  pro- 
duct  of  the   cosines    of  the   opposite   ex- 
j,        tr  ernes. 


Dem. — In  the  spliencal  triangle  BAC,  right 
angled  at  A,  taking  h,  c,  comp  B,  comp  C,  and 
comp  a  in  succession  as  middle  paits,  we  are  to 
prove  that 


sin  h  =  cos  (comp  a)  x  cos  (comp  B),  .  or    sin  b  =  sin  a  sin  B  ;  (1) 

sin  c  =  cos  (comp  a)  x  cos  (comp  C),    or    sin  c  =  sin  a  sin  C  ;  (3) 

sin  (comp  B)  =  cos  b  x  cos  (comp  C),    or  cos  B  =  cos  6  sin  C ;  (3) 

sin  (comp  C)  =  cos  c  x  cos  (comp  B),    or   cos  C  =  cos  c  sin  B  (4) 

sin  (comp  a)  =  cos  b  x  cos  c,  or  cos  a  =  cos  b  cos  c.  (5) 


To  demonstrate  these  relations,  let  O  be  tlie  centre  of  the  sphere,  and  draw 
the  radii  OA,  OB,  and  OC.  The  angles  BOC,  AOC,  and  AOB,  are  measured 
respectively  by  «,  b,  and  c,  the  sides  of  the  trianigle ;  hence  these  angles  at  the 
centie  and  their  measuring  ai'cs  may  be  used  interchangeably.    From  one  of 


NAPIER  S   RULES. 


77 


the  oblique  anj^les,  as  C,  let  fall  a  perpendicular  upon  the  radius  OA.  From 
the  foot  of  this  pei'pendicular  draw  DE  perpendicular  to  OB,  and  join  C  and  E. 
Now  CDE  is  a  right  angle  (Part  II.,  426),  CE  is  perpendicular  to  OB  (Part  II., 
S99),  and  DEC  is  equal  to  angle  B  of  the  triangle  (Part  II.,  55S). 

CD  =  sin  h,  OD  =  cos  b,  CE  =  sin  a,  and  OE  —  cos  a. 


(1) 


From  the  triangle  CDE,  right  angled  at  D,  we  have 

CD  =  CE  X  sin  CED,    or    sin  ft  =  sin  a  sin  B. 

Generalized,  (1)  becomes,  TTie  sirie  of  either  side  about  tlie  right  angle  —  the 
sine  of  the  hypotenuse  into  the  sine  of  the  angle  opposite  the  side.  Hence,  from 
analogy  to  (1),  we  may  write 

sin  c  =  sin  ct  sin  C. 


(3) 


Or  (2)  may  be  proved  in  the  same  manner  as  (1), 
by  letting  fall  from  B  a  perpendicular  upon  OA, 
from  its  foot  drawing  DE  perpendicular  to  OC, 
and  joining  E  and  B.  Then  BD  —  sin  c,  OD  = 
ccs  c,  BE  =  sin  a,  OE  =  cos  a,  and  angle  BED  = 
angle  C.    From  the  triangle  BDE,'we  have 

BD  =  BE  X  sin  BED,  or  sin  c  =  sin  a  sin  C.    (2) 

To  prove  (3),  we  have  from  the  triangle  CDE, 
Fig.  44, 

cos  CED  =  —  ,    or     cos  B 
CE 

But  from  triangle  OED,  right  angled  at  E,  ED  =  OD  x  sin  DOE  =  cos  5  sin  c  = 
[from  (3)],  cos  b  sin  a  sin  C.     Substituting  this  value  of  ED,  we  have 

_,       cos  6  sin  a  sin  C  ,    .    ^ 

cos  B  = r =  cos  b  sm  C.  (3) 

sma  ^ 

We  may  write  (4)  from  (3)  by  analogy,  as  (2)  was  from  (1) ;  or,  better,  let  the 
student  produce  it  from  Fig.  45,  as  (3)  was  produced  from  Fig.  44. 

Finally,  to  produce  (5),   consider  the  triangle  ODE,  in  either  figure,  right 
angled  at  E.    This  gives 


ED 

sin  6 


OE  =  OD  X  cos  DOE,    or    cos  a  =  cos  6  cos  c. 


(5) 


119,  JRtile  2.  JProjy* — -^^^  any  right  angled  splier- 
ical  triangle,  the  sine  of  the  middle  part  equals  the  i^o- 
duct  of  the  tangents  of  the  adjacent  extremes, 

Dem.— In  the  spherical  triangle  BAC,  right  angled  at  A,  taking 
by  c,  comp  B,  comp  C,  and  comp  a,  in  succession  as  middle  parts, 
we  are  to  prove  that, 


78 


SPHERICAL  TRIGONOMETRY. 


sin  b  =  tan  c  x  tan  (comp  C),    or  sin  b  =  tan  c  cot  C  ;  (1) 

sin  c  =  tan  &  x  tan  (comp  B),    or  sin  c  =  tan  6  cot  B ;  (2) 

sin  (comp  B)  =  tan  c  x  tan  (comp  a),    or  cos  B  =  cot «  tan  c  ;  (3) 

sin  (comp  C)  =  tan  b  x  tan  (comp  a),    or  cos  C  =  cot  a  tan  b  ;  (4) 

sin  (comp  a)  =  tan  (comp  B)  x  tan  (comp  C),  or  cos  a  =  cot  B  cot  C-  (5) 


Taking  theformiihv  of  Rule  1st,  and 
ting  the  value  of  each  factor  as  found 
write  tlie  followius^ 


sin  &  =  sin  a  sin  B  = 
sin  c  =  sin  a  sin  C  = 
cos  B  =  cos  6  sin  C  = 
cos  C  =  cos  c  sin  B  = 
cos  a  =  cosb  cos  c  = 


in  the  second  member  of  each  siibstitu- 
in  some  other  of  the  set,  we  readily 


sin  c  cos  C 

sine 
cose 

cosC 

=  tan  c  cot  C  ; 

(1) 

sin  C  cos  c 

sin  C 

sin  b  cos  B 

sin  b 

cos  B 
sinB 

=  tan  b  cot  B  ; 

(2) 

sin  B  cos  b 

~  cosb 

cos  a  sin  c 
cos  c  sm  a 

cos  a 

=  -. X 

sin  a 

sin  c 

COSC 

=  cot  a  tan  c  ; 

(3) 

CCS  a  sin  b 

cos  bsina' 

cos  a. 

=  -; —  *x 

sma 

sin  b 
cosb 

=  cot  a  tan  5  ; 

(4) 

cos  B  cos  C 
sin  C  sin  B 

cosB 
&m  B 

cosC 
sinC 

=  cot  B  cot  C. 

(5) 

Q.  E. 

D. 


120.  Sen.  1. 


Fig.  47. 


-It  will  be  a  good  exercise  for  the  student  to  demonstrate 

Rule  2d  from  the  an- 
nexed figures,  as  Rule  1st 
was  from  Figs.  44  and  4o. 
The  5th  is  not  as  readil}^ 
obtained  from  the  figure 
as  the  others.  The  stu- 
dent may  trace  the  fol- 
lowing relations,  some  .in 
one  figure,  and  some  in 
Fig.  4S.  the  Other. 


Cos  «  =  gi  =  '-^  =  cos  b  COS  c.    But,  cot  C  =  ^  =  '^\  and  cot  B  =  §5 
OE       sec  c  AE       tauc  AE 

s^^  '^        T.  ^  _,     ,  ^      sin  5  sin  e  ^  . «      .  /^ 

=  ^ — ;  ;  whence,  cot  B  cot  C  = r =  cos  b  cos  c.    /.  cos  a  =  cot  a  cot  C. 

tan  6  tan  6  tan  c 

121,  ScH.  2. — It  is  of  much  importance,  especially  for  the  purposes  of 
Spherical  Astronomy,  that  the  student  observe  that  the  relations  expressed  in 
the  above  formulae,  and  in  fact  all  the  relations  between  the  sides  and  angles  of 
spherical  triangles,  are  also  the  relations  between  the  facial  and  diedral  angles 
of  triedrals.  Thus,  if  a,  b,  and  c  represent  the  facial  angles,  and  A,  B,  and  C  the 
opposite  diedrals,  all  these  relations  can  be  established,  and  in  exactly  the  same 
manner  as  above,  without  any  allusion  to  the  ep/ierical  triangle.  [The  student 
should  do  it.l 


DETERMINATION   OF   SPECIES.  79 


DETERMINATION    OF    SPECIES. 


122.  In  the  solution  of  spherical  triangles  the  determination  of 
the  species  of  a  part  sought  becomes  of  essential  importance,  since 
any  part  of  such  a  triangle  may  have  any  value  l^etween  0°  and  180°. 
Hence,  when  we  have  learned  the  numerical  value  of  any  function  of 
a  part,  we  have  yet  to  determine  whether  the  part  itself  is  less  or 
greater  than  90°,  i.  e.,  the  species  of  the  part.  This  may  always  be 
effected  by  some  one  of  the  following  propositions. 


123,  JProp, — If  the  part  sought  is  found  in  terms  of  its  cos, 
tan,  or  cot,  its  S2)ecies  can  he  determined  hy  the  algebraic  signs  of  the 
functions  in  the  formula  used. 

Dem. — In  each  oi  i\ve  formulm  arising  from  Ihe  application  of  Napier's  rules, 
there  are  three  functions,  the  arcs  corresponding  to  two  of  which  are  always 
known,  hence  the  algebraic  signs  of  their  functions  are  known,  and  the  signs 
of  these  two  determine  the  sign  of  the  third  or  unknown  function.  Now,  when 
a  cos,  tan,  or  cot  is  +  ,  the  corresponding  angle  is  less  than  90°  (if  less  than 
180°) ;  and  when  one  of  these  functions  is  — ,  the  corresponding  angle  is  greater 
than  90° ;  i.  e.,  in  a  triangle,  it  is  between  90°  and  180°. 


124.  When  -the  part  sought  is  found  in  terms  of  its  sine,  the 
species  cannot  be  determined  by  the  signs  of  the  formula,  since  the 
part  being  less  than  180°  its  sine  is  always  +.  The  three  following 
propositions  dispose  of  such  cases. 


125,  ^rop, — An  ohlique  angle  of  a  right  angled  spherical  tri- 
angle and  its  opposite  side  are  always  of  the  same  species. 

'  Dem.— From  Napier's  first  rule  we  have,  cos  B  =  cos  J  sin  C.  But  sin  C  is 
necessarily  +  ;  therefore,  cos  B  and  cos  h  always  have  the  same  sign,  and  B  and 
h  are  of  the  same  species.  In  like  Dianner,  we  see  from  cos  C  =  cos  c  sin  B,  that 
C  and  c  are  of  the  same  speaies.    q.  e.  d. 


126,  JProp, —  When  the  hypotenuse  of  a  right  angled  spherical 
iriangle  is  less  than  90°,  the  other  two  sides  {and  consequently  the 


80  SPHERICAL    TRIGONOMETRY. 

ohUque  angles)  are  of  the  same  species  with  each  other.  But  tvhen  the 
hypotenuse  is  greater  than  90'^,  tJie  other  two  sides  {and  consequently 
the  oblique  angles)  are  of  different  species  from  each  other, 

Dem. — From  Napier's  first  rule  we  have,  cos  a  —  cos  h  cos  e.  Now,  if 
a  <  90%  cos  a  is  +  ;  lience  cos  h  and  cos  c  must  have  like  si^^s,  and  b  and  c  be 
both  less  or  both  greater  than  90°  But  if  a  >  90'  (and  less  than  180%  as  it  is), 
cosrt  is  —  ;  hence,  cos  b  and  cos  c  must  have  different  signs,  and  b  and  c  be  one 
greater  and  the  other  less  than  90%  Finally,  since  the  oblique  angles  are  of  the 
same  species  as  their  opposite  sides,  they  are  of  the  same  species  with  each  other 
when  a  <  90%  and  of  different  species  from  each  other  when  a  >  90% 


127.  I^vOjy* — TT72f?i  a  side  and  its  opposite  angle  are  given  in  a 
right  angled  sp^herical  triangle,  there  is  xo  solution  if  the  sine  of  the 
side  is  greater  than  the  si7ie  of  its  opposite  angle;  there  is  one 
solution  and  the  triangle  is  hi-rectangular  if  the  sine  of  the  side 
equals  the  sine  of  its  opposite  angle ;  and  there  are  two  solutions  if 
the  sine  oftlie  side  is  less  than  the  sine  of  its  opposite  angle. 

Dem. — We  have  sin  5  =  sin  a  sin  B,  or  sin  a  =  -: — = .    Now,  sin  6  >  sin  B 

sm  B 

makes  sin  a>  1,  which  is  manifestly  impossible.     Sin&  =  sin  B  makes  6  =  B, 

since  they  are  of  the  same  species.     But  when  the  arc  included  by  tJie  sides  of 

an  oblique  angle  of  a  right  angled  spherical  triangle  is  equal  to  the  angle,  the 

vertex  of  the  angle  is  the  pole  of  the  arc.     Hence,  in  this  case  the  other  sides  of 

the  triangle  are  each  90%    Finally,  if  sin  b  <  sin  B,    a    has    two    values,  one 

^eater  and  the  other  less  than  90°.    Hence  there  are  two  triangles. 

128,  ScH. — These  relations  between  an  angle  and  its  opposite  side  may  be 
observed  directly  from  a  figure.     When  B  <  90%  the  measure  of  it,  that  is 

6  =  B,  is  tlie  greatest  included  perpendicular  which 
can  be  drawn  to  one  side  of  the  lune  BAB'C.  Hence, 
in  tliis  case,  b  cannot  exceed  B,  which  implies  that 
sin  b  cannot  be  >  sin  B,  as  when  the  arcs  are  less 
than  90',  the  greater  arc  has  the  greater  sine.  If 
sin  6  =  sin  B,  b  =  B,  and  BA  =  BC  =  90%  and  the 
side  b  can  occupy  but  one  position  in  the  lime,  thus 
giving  rise  to  but  one  tiiangle  BAC,  which  satisfies 
the  conditions  (or  two  equal  triangles  BAC  and 
B'AC).  If  6  <  B,  which,  when  B  is  less  than  90° 
unplies  that  sin  b  <  sin  B,  the  side  can  occupy  two  positions  m  the  lime,  b'  and 
b"  giving  rise  to  two  triangles,  BA'C  and  BA"C'%  both  of  which  fulfill  the 
conditions. 


EXERCISES  IN  SOLVING  RIGHT  ANGLED  TRIANGLES. 


81 


Again,  when  B  >  90",  the  measure  of  it,  i.  e.,  6  =  B, 
perpendicuhir  that  can  be  drawn  to  one  side  of  the 
lune.  Hence,  in  this  case  we  cannot  have  &  <  B, 
which  imphes  that  sin  h  cannot  be  >  sin  B,  since  the 
greater  arc  has  the  less  sine.  If  sin  b  —  sin  B, 
J  =  B,  and  BA  =  BC  r=  90°,  and  the  side  h  can  oc- 
cupy but  one  position  in  the  lune,  thus  giving  rise 
to  but  one  triangle  BAC,  which  satisfies  the  conditions 
(or  two  equal  triangles  BAC  and  B'AC).  If  6  >  B, 
which  implies  that  sin  h  <  sin  B,  the  side  b  can 
occupy  two  positions  in  the  lune,  as  b'  and  b", 
thus  giving  rise  to  two  triangles  BA'C,  and  BA"C' 
conditions. 


is  the  least  included 
C 


A' 

Fig.  50. 

both  of  which  satisfy  the 


EXERCISES. 

1.  In  a  right  angled  spherical  triangle  BAC,  A  being  the  right  angle, 
B  =  80°  40',  and  a  =  105°  34',  to  project  the  triangle  and  compute 
the  other  parts. 

Projection.* — Projecting  the  triangle  upon  the  plane  of  the  side  c  (102), 
we  have,  BCA,  Fig.  51.     [The  student  should  give 
the  process.] 

Solution.— It  is  immaterial  which  of  the  re- 
quired parts  we  seek  first.  We  will  seek  c.  Now 
the  three  circular  parts  under  consideration  are  c, 
comp«,  and  comp  B.  Comp  B  is  middle  part, 
and  the  extremes  are  adjacent ;  hence,  by  Napier's 
second  rule  we  have, 

cos  B  =  tan  c  cota. 
cos  B       cos  80°  40' 


or  tan  c=  _        ,^^,  ,,,,. 

cot  a      cot  lOo   d4 

Now  cos  80°  40'  is  + ,  and  cot  105°  34'  is 


Fig.  51. 
Therefore  tan  c  is  — ,  and  c  >  90°. 


Computing  by  logarithms, 

log  cos  80°  40'  =    9.209992 
-  log  cot  105°  34'  =    9.444947 

=  log  tan  c  =    1.765045 

Add  10  for  tab.  tan       10. 


9.765045.     .-.  c  =  149°  47'  37' 


*  It  is  recommendecl  that  the  projection  be  given  ahvays  before  the  trigonometrical  eolution. 
It  is  an  excellent  exercise,  and  gives  clearness  of  perception. 

6 


82  SPHERICAL  TRIGONOMETRY. 

To  find  b,  sin  b  =  sin  a  sin  B  =  siu  105'  34'  x  sin  80°  40'.  This  makes  b 
known  by  means  of  its  sine,  whence  the  signs  of  the  formula  do  not  determine 
the  species  of  6.  But  b  is  of  the  same  species  as  B  {124)^  and  therefore  less 
than  90°. 


Computing  by  logarithms, 

log  sin  105°  34'  =  9.983770 
+  log  sin  80°  40'    =  9.994212 

Deducting  10    =  9.977983  =  log  sin  5.   .-.  b  =  71°  54'  33". 


rr      ^     1   r^  *  B        ^  ^  ^ «  cos  «  COS  105°  34'       __ 

To  find  C,  COS  a  =  cot  B  cot  C,  or  cotC  =  — — ^   =   —  .    Whence 

cot  C  is  -,  and  C  >  90°. 

Computing  by  logarithms, 

log  cos  105°  34'  =    9.428717 
-  log  cot  80°  40'    =    9.215780 

Adding  10  =  10.212937  =  log  cot  C.  .'.  C  =  148°  30'  54". 

SCH. — It  is  expedient  to  find  each  part  directly  from  the  parts  given  in  the 
example,  in  order  that  an  error  in  finding  one  may  not  extend  itself  through 
the  -whole  solution. 


2.  Given  a  =  86°  51',  and  B  =  18°  03'  32",  to  project  the  triangle 
and  compute  the  other  parts. 

c  =  86°  41'  14",  b  =  18°  01'  50",  C  =  88°  58'  25". 

3.  Given  b  =  155°  27'  54",  and  c  =  29°  46'  08",  to  project  the 
triangle  and  compute  the  other  parts.  See 
Fig.  52. 

a  =  142°  09'  13",  C  =  54°  01'  16", 
B  =  137°  24'  21". 

4.  Given  c  =  73°  41'  35",  and  b  = 
99°  17'  33",  to  project  the  triangle  and' 
compute  the  other  parts. 

C  =  73°  54'  46",    b  =  99°  40'  30", 
Fxa.  32.  «  =  9^°  42'  17". 

5.  Given  B  =  47°  13'  43",  and  c  =.  126°  40'  24",  to  project  the 
triangle  and  compute  the  other  parts. 

b  =  32"  08'  56",  a  =  133°  32'  26",  c  =  144°  27'  03". 


EXERCISES  IN  SOLVING  EIGHT  ANGLED  TRIANGLES, 


83 


Projection. — In  order  to  project  this  case,  i. «., 
when  the  two  oblique  angles  are  given  {105),  it  is 
most  convenient  to  compute  the  base  before  pro- 
jecting. It  is  also  expedient,  when  two  angles  are 
given,  to  project  the  larger  at  a  point  in  the  cir- 
cumference of  the  primitive  circle,  as  at  C,  espe- 
cially if  the  smaller  be  quite  small.  In  this  case, 
projecting  the  angle  C  at  C,  Fig.  53,  conceive  BA  as 
drawn  through  P  (or,  if  desired,  sketch  it  hj-po- 
thetically),  and  then  compute  b,  from  the  relation 

cos  B 


cos  B  =  sin  C  cos  5,  or  cos  h  •■= 
found  h 


.     ^.    Having 
sm  C  ^ 

32°  08'  56",  take  CA  =  b,  and  draw  AB  through 


Fig.  53. 


P. 


6.  Given  B  =  100°,  and  b  =  112°,  to  project  the  triangle  and  com- 
pute the  other  parts. 

Projection.— See  Fig.  54. 

Numerical  Solution.— r<?  find  c,  we  have 

sin  c  =  tan  6  cot  B  =  tan  112°  cot  100\ 

Computing  by  logarithms, 

log  tan  112°  =  10.893590 
+  log  cot  100°  — -    9.246319 

Rejecting  10  =    9.639909  =  log  sin  c. 

.'.  c  =  25°  52'  33".4,  or  154°  07'  26".6,  i.  e.,  BA,  or  BA'. 


Fig.  54. 


To  find  a,  we  have 

sin  &  =  sin  a  sin  B ;  whence  sin  a  = 


sin  b 


sin  112' 


sin  B 


sm  lUO 


Computing  by  logarithms, 

log  sin  112°  =  9.967166 
-  log  sin  100°  =  9.993351 

Adding    10   =  9.973815  =  log  sin  a.  .-.  <i  =  70°18' 10".7,or  109' 41' 49".3, 
t.  e.,  BC,  or  EC. 


To  find  C,  we  have 

cos  B  =  cos  5  sin  C  ;  whence  sin  C  = 


cos  B      cos  100° 


cos  b      cos  113' 


Computing  by  logarithms, 

log  cos  100°  =  9.239670 
-  log  cos  112°  =  9.573575 

Adding    10   =  9.006095  =  log  sin  C.   .-.  C  =  27°  36'  oS".S,  or  152°  23'  01".2, 
t>.,  BCA,  Or  BC'A'. 


Si  SPHERICAL  TRIGOXOMETBY. 

Thus  -we  see  that  each  of  the  two  triangles  BCA  and  BC'A'  fulfills  the  con- 
ditions of  the  problem. 

T.  Given  one  side  of  a  right  angled  si^herical  triangle  160°,  and  the 
opposite  angle  150°,  to  project  the  triangle  and  compute  the  other 
parts. 

Jtesidts. — There  are  two  triangles.  The  other  sides  of  the  first 
are  136°  50'  23",  and  39°  01'  51";  and  the  angle  opposite  the  latter 
side  is  67°  09'  43".  The  corres^Donding  j^arts  of  the  other  triangle 
are  13°  09'  37",  110°  55'  09",  and  112°  50'  17". 

8.  In  the  spherical  triangle  DEF,  right  angled  at  E,  given  an  oblique 
angle  58°,  and  the  side  oj^posite  61°,  to  project  the  triangle  and  com- 
pute the  other  parts. 

9.  In  a  right  angled  s]oherical  triangle  given  an  oblique  angle 
165°,  and  the  o^Dposite  side  112°,  to  project  the  triangle  and  com- 
pute the  other  parts. 

10.  In  a  right  angled  spherical  triangle  given  one  side  65°  23'  12", 
and  the  opposite  angle  6b°  23'  12",  to  project  the  triangle  and  com- 
pute the  other  parts. 

11.  Given  c  =  60°  47'  24".3,  B  =  57°  16'  20".2,  and  A  =  90°,  to 
project  and  compute. 

a  =  68°  56'  28".9,  c  =  51°  32'  32".l,  and  b  =  51°  43'  36".l. 

12.  Given  c  =  116°,  b  =  16°,  and  the  included  angle  90°,  to  pro- 
ject and  compute. 


QUADRANTAL  TKLLXGLES. 

129.  A  QiiadranUil  Triangle  is  a  spherical  triangle  one 
of  whose  sides  is  a  quadrant,  or  90°.  Such  a  triangle  is  readily 
solved  by  passing  to  its  polar,  solving  it,  and  then  passing  back. 
The  polar  triangle  to  a  quadrantal  triangle,  being  right  angled,  is 
solved  by  Napier's  rules. 

Ex.  1.  Given  a  =  90°,  B  =  75°  42',  and  c  =  18°  37',  to  compute 
the  other  parts. 

Sug's. — Representing  the  supplemental  parts  of  the  polar  triangle  by  A',  B', 
C,  a',  b',  and  c',  we  have  A'  =  ISO*  -  a  =  90',  b'  =  180'  -  B  =  101°  18',  and 


OF    OBLIQUE   ANGLED   SPHERICAL  TEIANGLES. 


85 


C  =  180°  -  c  =  1G1°  23',  from  which  to  find  B',  a',  and  c'.  This  being  rii^ht 
angled,  we  find,  by  applying  Napier's  rules,  B'  =  94°  31'  21",  a'  =  76°  25'  11", 
and  c'  =  161°  55'  20".  Hence  in  the  primitive  triangle  we  have  b  =  180°  -  B* 
=  85°  28'  39",  A  =  180°  -  a'  =  103°  34'  49",  and  C  =  180°  -  c' =  18°  04'  40 ". 

Ex.  2.  Given  a  =  90°,  C  =  42°  10',  and  A  =  115°  20',  to  find  the 
other  parts. 

B  =  54°  44'  24",  b  =  64°  30'  40",  c  =  47°  57'  47". 


SECTION  II, 


OF  OBLIQUE  ANGLED  SPHERICAL  TRLINGLES. 

±30,  All  cases  of  oblique  angled  spherical  triangles  can  be  solved 
by  Napier's  rules  and  the  following  proposition. 

1.S1,  ^rop, — In  any  spherical  triangle,  if  a  perpeyidicular  be 
let  fall  fiwn  either  vertex  upon  the  opposite  side  or  side  produced, 
the  tangent  of  half  the  siim  of  the  segments^  of  that  side  is  to  the 
tangent  of  half  the  sum  of  the  other  tivo  sides,  as  the  tangent  of  half 
the  difference  of  those  sides  is  to  the  tangent  of  half  the  difference 
of  the  segments. 

Dem. — In  the  triangle  BAC  let  fall  the  perpen- 
dicular p,  from  C  upon  the  opposite  side.  Let 
BD  =  8,  and  DA  =  s'.  By  Napier's  first  rule, 
cos  a  =  cos  p  cos  s,  and   cos  b  =  cos  p  cos  s'. 

Dividing  the  former  by  the  latter, =  — — , ; 

^  •'  cos  b      cos  s  ' 

whence,  by  composition  and  division, 

cos  b  —  cos  a      cos  s'  —  cos  s 


cos  a  +  cos  b      cos  s  +  cos  s'' 


Fig.  55. 


But  by  (6i), 


and 


cos  b  —  cos  a 
cos  a  +  cos  b 

cos  s'  —  cos 


=  tan  i{a  +  b)  tan  i  (a  —  J), 


cos  8   +   COS  8 


>  =  tan  i  {s  +  s')  tan  ^{s  —  s'), 


*  When  the  perpendicular  falls  without  the  hase,  a?  in  Fig.  50,  this  term  is  to  be  understood 
as  meaning;  the  distances  from  the  foot  of  the  perpendicular  to  each  extremity  of  the  base,  as 
BD  and  AD-  This,  in  fact,  is  the  general  statement— applying  as  well  to  the  case  when  the 
perpendicular  falls  on  the  base. 


86 


SPHERICAL  TPilGOXOMETEY. 


.-.    tan  i {a  +  b)  tan  |(a  —  5)  =  tan  i (s  +  s')  tan  i («  —  «') ; 

tan  i{s  +  s')  :  tan  ^  {a  +  b)  : :  tan  ^{a  —  b)  :  tan  ^{s  —  «').    q.  e.  D. 

132,  ScH.  1. — Since  from  a  point  in  the  surface 
of  a  hemisphere  two  perpendiculars  can  always 
be  di-awn  to  the  circumference  of  the  great  cii'cle 
which  forms  its  base,  and  since  the  feet  of  thesq 
perpendiculars  are  180°  apart,  and  no  side  of  a 
spherical  triangle  can  equal  180°,  the  foot  of  one 
perpendicular  will  always  fall  within  the  base  or 
upon  one  extremity  of  it,  and  the  other  without 
the  base ;  or  both  will  fall  without  the  base.  K 
we  take  the  foot  of  the  pei-pendicular  which  falls 
within  the  base,  or  the  nearer  one  when  both  fall 
without,  the  sum  of  the  distances  from  the  foot  of 
the  perpendicular  to  the  extremities  of  the  base  is  always  less  than  180°^ 
t.  €.,  s  +  s'  <  180°.  When  the  perpendicular  falls  within,  8  +  s'  makes  up  one 
side  of  the  triangle,  and  hence  is  less  than  180°.  If  both  perpendiculars  fall 
without,  let  D,  Fi{/.  56,  be  the  foot  of  the  nearer  one.  Now  DB  +  BD'  =  180° ; 
but  by  hypothesis  DA  <  BD',  .-.  DB  h-  DA  <  180°.  When  DA  =  BD',  DB  +  DA 
=  180°. 

133.  ScH,  2. — As  in  spherical  triangles  the  greater  segment  is  not  always 
adjacent  to  the  greater  side,  it  becomes  necessar}'  to  determine  the  position  of 
the  segments.    This  can  be  done  by  the  signs  of  the  proportion 

tan  i  {s  —  s). 


Fig.  56. 


tan  i  (s  +  s')  :  tan  ^  {a  +  b)  :  :  tan  i  {a  —  b) 


1st.  Tan  ^{s  +  «')  is  always  + ,  since,  if  D  falls  in  the  base,  8  +  s'  <  180° ; 
and  if  D  falls  without,  by  taking  the  nearer  perpendicular,  «  +  &'  is  still  <  180° 
(132).    .:  H«  +  «')  <  90°,  and  tan  H«  +  *')  is  + . 

2d.  When  a  +  b  <  180°,  tan  i  (a  +  5)  is  +  ;  and  when  a  +  b>  180°. 
tan  i  {a  +  b)  is  — . 

3d.  When  a  >  b,  a  —  b  is  a.  positive  arc  less  than  180°,  hence  tan  ^{a  —  b) 
is  +  ;  and  when  a  <b,  (a  —  6)  is  a  negative  arc  and  less  than  180°,  hence 
tan  i  (^  —  ^)  is  — . 

4th.  The  signs  of  these  terms  being  determined,  that  of  tan  i  (s  —  s')  becomes 
known.  Now,  as  ^{s  —  s')  cannot  be  numerically  greater  than  90°,  tan  i {s  —  s) 
is  +  when  s  >  s\  and  —  when  s  <  s'. 

5th.  When  8  +  8'  =  180°,  tan  i{s  +  s')=  oc.  Now  as  a  -  6  <  180°,  tan  i  {a-b) 
cannot  be  oc,  nor  can  tan  ^  {s  —  s')  =  0  when  the  perpendicular  falls  without. 
Hence  to  make  the  proportion  possible, tan  ^{a  +  b)  must  be  cc,or  a  +  b=  180°. 
In  this  case  we  project  on  the  plane  of  a  or  b.  If:  a  +  b  =  180",  aad  a  +  G  =  180°, 
we  project  on  the  plane  of  a.  U  a  +  b  =  180°,  a  +  c  =  180°,  and  b  +  c  =  180°,  the 
tiiangle  is  trirectangular. 

134,  ScH.  3. — If  either  segment  is  greater  than  the  whole  base,  the  perpen- 
dicular falls  without  the  triangle.  In  this  case  the  shorter  segment  lies  in  an 
opposite  direction  from  its  angle  to  that  considered  in  the  demonstration,  and 
hence  is  to  be  considered  —  ;  and  the  algebraic  sum  of  the  segments  is  still 
equal  to  the  side  upon  which  the  perpendicular  is  let  fall. 


OBLIQUE  ANGLED  TRIANGLES  SOLVED  BY  NAPIER's  RULES.   87 

135*  Prop. — In  a  spherical  triangle,  the  sines  of  the  sides  are 
to  each  other  as  the  sines  of  their  0])posite  angles. 

Dem. — By  Napier's  first  rule  we  have  from  either  Fig.  55  or  Fig.  56, 

sin  p  =  sin  a  sin  B,    and    sin  ^  =  sin  5  sin  A. 

.'.  sin  a  sin  B  =  sin  h  sin  A,    or    sin  a  :  sin  5  :  :  sin  A  :  sin  B.     q.  e.  d. 

ScH. — This  proposition  is  not  introduced  here  because  it  is  necessary  for  the 
solution  of  spherical  triangles,  but  because  of  its  essential  importance.  It  is 
often  convenient  to  use  it  in  the  solution  of  a  triangle,  but  never  necessary,  as 
will  appear  hereafter.  It  affords  a  ready  metlwd  of  determining  a  part  opposite 
a  given  part,  provided  the  species  of  the  part  he  determined  by  other  considerations. 


SOLUTIOX  OF  OBLIQUE  ANGLED  SPHERICAL  TRIANGLES  BY 
NAPIER'S  RULES  FOR  RIGHT  ANGLED  SPHERICAL  TRIANGLES. 

130,  The  examples  which  arise  in  the  solution  of  oblique  angled 
spherical  triangles  are  all  comprised  under  the  three  following 
problems,  each  of  which  consists  of  two  cases : — 

1.  When  the  given  parts  are  all  adjacent  to  each  other. 

2.  When  two  of  the  given  parts  are  adjacent  and  one  separate. 

3.  When  the  given  parts  are  all  separate  from  each  other. 


137,  J^voh*  1. — Given  three  adjacent  parts  of  an  ohlique  angled 
spherical  triangle,  to  solve  the  triangle. 

Case  1st. — Given  tiuo  sides  and  the  included  angle. 
Solution. — Project  the  triangle  on  the  plane  of  one  oftlie  given  sides  {106), 
and  let  fall  a  perpendicular  from  the  angle  opposite  upon  this  side  or  upon 


Fig.  58. 

this    side, produced,  as  the  case  may  be.     There  are  thus  formed   two  right 
angled  triangles,  as  BDC  and  DC  A,  each  of  which  can  be   solved  by  Napier's 


88 


SPHERICAL  TRIGONOMETRY. 


rules,  by  first  solving  the  one  containing  the  given  angle.  Thus,  in  the  triangle 
BDC  right  angled  at  D,  a  and  B  are  supposed  known ;  whence  CD,  BD,  and  the 
angle  BCD,  can  be  computed.  As  BA  =  c  is  known,  the  segment  DA  can 
be  found,  it  being  the  difiference  between  c  and  the  arc  BD.  When  the  solu- 
tion of  tliis  triangle  gives  BD  >  <•,  it  is  evident  that  the  perpendicular  falls 
without  tlie  triangle,  which  will  agree  with  the  projection.  Passing  to  the 
triangle  ADC,  right  angled  at  D,  we  now  know  CD  and  DA ;  whence  the 
other  parts  can  be  found.  Finally,  the  angle  BCA  of  the  required  triangle  = 
BCD  +  DCA  when  the  perpendicular  falls  witliin  the  triangle,  and  BCD  —  DCA 
when  the  perpendicular  falls  without. 

Case  2d. — Given  tico  angles  and  the  included  side. 

Solution. — The  solution  of  this  case  is  effected  by  passing  to  the  polar 
triangle,  projecting  and  solving  it  by  Case  1st,  and  then  passing  back. 

138.  ScH. — A  slight  saving  of  labor  is  effected  by  usin^  {135)  in  the  solu- 
tion. Thus,  in  the  triangle  BCD,  compute  CD  and  BD  as  before,  and  (not  com- 
puting angle  BCD)  then  passing  to  the  triangle  DCA,  compute  h  and  A.  Finally, 
compute  C  (the  entu-e  angle)  from  the  proportion 

sin  6  :  sin  c  : :  sin  B  :  sin  C. 


139.  JProb,  2.-^1  an  oblique  angled  spherical  triangle,  given 
fivo  2)arts  adjacent  to  each  other  and  one  separated  from  both  of  them, 
to  solve  the  triangle. 

Case  1st. — Given  two  sides  and  an  angle  opposite  one  of  them. 

Solution. — Project  the  triangle  on  the  plane  of  the  unknown  side,  with  the 
given  angle  at  B ;  and  let  fall  the  perpendicular  from  the  angle  C  opposite  the 
unknown  side.    Compute  the  tiiangle  BDC.    Having  computed  this  triangle. 


Fig,  60. 

compare  the  side  opposite  the  given  angle,  as  5,  with  the  perpendicular  and 
the  arcs  BC  and  CB',  i.  e,,  with  ^,  a,  and  180°  —a.    If  b  =  p  there    is  but 


OBLIQUE  ANGLED  TRL^GLES  SOLVED  BY  NAPIER's  RLXES.        89 

one  solution  and  the  triangle  is  right  angled,  A  falling  at  D.  If  b  is  inter- 
mediate in  value  between  p  and  both  a  and  180°  —  «,  it  can  occupy  two  positions 
as  in  Fig.  59,  and  there  are  two  solutions.  If  b  is  intermediate  in  value  between 
J)  and  only  one  of  the  arcs  a  or  180°  —  a,  there  is  but  one  solution.  When  B  <  90° 
the  perpendicular  is  less  than  any  oblique  arc;  hence  in  this  case,  if  Z>  <  j9,  there 
is  710  solution.  But  if  B  >  90°,  the  pei-pendicular  is  greater  than  the  obhque 
arcs ;  hence  in  this  case,  \ib>  p,  there  is  no  solution.  [These  results  should  be 
obtained  independently  of  the  results  given  by  the  projection,  and  one  be  made 
a  check  upon  the  other.]  The  solution  is  now  completed  by  computing  the 
parts  of  DCA,  and  adding  or  subtracting  the  segments  BD  and  AD,  and  the  angles 
BCD  and  A  CD,  as  the  case  may  require. 

Case  2d. — Given  tivo  angles  and  a  side  opposite  one  of  them. 

Solution. — ^Pass  to  the  polar  triangle ;  solve  it,  and  then  pass  back.* 

14:0,  ScH. — The  relation  established  in  {13o)  may  also  be  used  in  the 
solution  of  this  problem.  Thus,  having  projected  the  triangle,  computed 
j9,  and  determined  whether  there  are  one  or  two  solutions,  to  find  A,  when 
«,  by  and  B  are  given,  we  have,  sin  b  :  sin  a  :  :  sin  B  :  sin  A.  Then  computing 
the  third  side  c  (or  sides),  by  means  of  the  right  angled  triangles  BCD  and  DCA 
as  before,  we  may  use  the  proportion  {135)  to  find  the  angle  BCA  and  BCA'. 
But  the  use  of  this  proportion  gives  no  advantage  except  in  cases  in  which 
there  is  only  one  solution. 


14:1,  JPvoh.  3, — In  an  oblique  angled  sp)herical  triangle,  given 
three  parts  all  separated  from  each  other,  to  solve  the  triangle. 

Case  1st. —  Given  the  sides  to  find  the  angles. 

Solution. — Project  the  triangle  on  the  plane  of  one  of  its  sides,  as  c.    From 
the  proportion, 

tan|(s  4-  s')  :  tan  \{a  +  b)  :  :  tan  i{a  —  b)  :  tan  i{s—s'), 

*  This  case  can  be  projected  and  solved  in  a  manner 
altogether  similar  to  the  first,  without  passing  to  the 
polar  triangle.  Thus,  let  B,  «i  and  A  be  the  given  parts. 
]^roject  the  triangle  on  the  plane  of  c,  as  in  the  figure. 
Project  B  in  the  usual  way,  and  make  BC  =  the  projec- 
tion of  a.  Through  C  draw  DD',  and  make  BDO  =  the 
projection  of  A-  Drawing  the  small  circle  with  radius 
PC,  draw  diameters  through  the  intersections  0  and  Q\ 
Then  will  A  and  A'  ^c  the  vertices  of  the  triangle  re- 
quired. The  student  may  prove  that  the  figures  POD 
and  PCA  are  equal,  and  also  PQ'D  and  PCA',  and 
hence  that  angle  BDO  =  A  =  A'. 


90 


SPHERICAL  TRIGONOMETRY. 


Fig.  62. 


find  i{s  —  s').  Then  half  the  sum  of  the  segments 
+  half  the  difference  gives  the  greater  segment,  and 
half  the  sum  —  half  the  difference  gives  the  less. 
Determine  from  the  signs  of  the  terms  whether  s  is 
greater  or  less  than  s'  :  and  also  determine  whether 
the  perpendicular  lies  wiUiin  or  without  the  tri- 
angle {134).  Observe  that  these  results  con-espond 
to  those  given  by  the  projection.  Finally,  in  each 
of  the  two  right  angled  triangles  BCD  and  DCA, 
there  are  two  sides  given ;  whence  the  angles  can 
be  found  by  Napier's  rules.  If  the  perpendicular 
falls  wiUiin,  C  =  BCD  +  DCA,  and  A,  of  the  re- 
quired triangle  =  DAC.  If  the  perpendicular  falls 
without,  C  =  BCD  —  DCA,  and  A  of  the  triangle 
=  180'  -  DAC. 


Case  2d. 
sides. 


■Given  the  angles    to  find   the 


Solution. — ^Pass  to  the  polar  triangle ;  solve  it, 
and  then  pass  back. 

14:2,  ScH.— Here,  again,  {135)  affords  a  slightly 
more  expeditious  solution.     Having  projected  the 
triangle,  found  and  located  the  segments,  and  com- 
puted one  angle,  as  B,  by  the  methods  given  above,  the  other  angles  may  be 
found  from  the  proportions, 

sin  b  :  sin  a  :  :  sin  B   :  sin  A, 
and  sin  Z>  :  sin  c   :  :  sin  B  :  sin  C. 


EXERCISES. 


1.  Given  l  =  120°  30'  30",  c 
to  project  and.  solve  the  triangle. 


70°  20'  20",  and  A  =  50°  10'  10", 


Projection. — See  Fig.  64. 

Trigonometrical  Solution.— 1st.  To  sohf  tlie 
triangle  ABD,  in  which  the  two  known  parts  Qre 
situated. 

(a)  To  find  p,  sin  p  =  smc  sin  A. 

log  sin  70'  20'  20"  =  9.973913 
-f-  log  sin  50°  10'  10  '  =  9.885329 

Rejecting  10  =  9.859241  =  log  sin  p 
.*.  p  =  46°  19'  01",  tlie  species  being  determined 
by  the  opposite  angle  {125).     [Observe  that  the  re- 
sult corresponds  with  the  projection]. 


OBLIQUE  ANGLED  TRIANGLES  SOLVED  BY  NAPIER's  EULES.   91 

(ft)  To  find  AD,  cos  A  =  cote  tan  AD,  or  tan  AD  =  -4-  . 
^  '  '  '  cot  c 

log  cos  50°  10'  10"  =    9.806532 

-  log  cot  70°  20'  20"  =    9.553016 

Adding  10  =  10.253516  =  log  tan  AD.    .-.  AD  =  60°  50'  49", 
the  species  being  determined  by  the  signs  of  the  formula. 

(c)  To  find  angle  ABD,  cos  c  =:  cot  A  cot  ABD,  or  cot  ABD  =:  — —. 

log  cos  70°  20'  20"  =:  9.526929 

-  log  cot  50°  10'  10"  =  9.921204 

Adding  10  =  9.605725  =  log  cot  ABD.     .-.  ABD  =  08°  01'  53" 
the  species  being  determined  by  the  signs  of  the  formula. 

2d.  To  solve  the  triangle  DBC. 

[a)  To  find  DC.    Since  AD  <  5,  the  foot  of  the  perpendicular  falls  in  the  base, 
and  DC  =  AC  -  AD  =  J  -  AD  =  120°  30'  30"  -  60°  50'  49"  =  59°  39'  41". 

{b)  To  find  a,  cos  a  =  cosp  cos  DC 

log  cos  46°  19'  01"  =  9.839270 
+  log  cos  59°  39'  41"  =  9.703386 

Rejecting  10  =  9.542656  =  log  cos  a.    /.  a  =  69°  34'  56",  the 
species  being  determined  by  the  signs  of  the  formula. 

sin  T) 

(c)  To  find  C,  sin  j;  =  sin  a  sin  C,  or  sin  C  =  — — . 

''  sm  a 

log  sin  p  =  9.859241 

-  log  sm  69°  34'  56"  =  9.971820 

Adding  10  =  9.887421  =  log  sin  C.     .'.  C  =  50°  30'  08",  the 

species  being  determined  by  the  side  opposite. 

{d)  To  find  angle  DBC,  smp  =  tan  DC  cot  DBC,  or    cot  DBC  =  -^~. 

tan  uC 

log  sin  2?  =    9.859241 

-  log  tan  59°  39'  41"  =  10.232653 

Adding  10  =    9.626588  =  log  cot  DBC.    .*.  DBC  =  67°  03'  36", 

the  species  being  determined  by  the  signs  of  the  formula. 

Finally,  B  =  ABD  +  DBC  =  68°  01'  53"  +  67°  03'  36"  =  135°  05'  29". 


ScH.— We  might  have  omitted  the  computation  of  angle  ABD  in  the  first 
part,  and  DBC  in  the  second,  and  have  found  instead  the  entire  angle  B  from 
sin  a  :  sin6  :  :  sin  A  :  sin  B.  To  compute  this  requires  the  looking  out  of  but 
two  logarithms,  since  sin  a  is  given  in  the  second  part  (c),  and  sin  A  in  the  first 
part  {a). 

2.  Given  a  =  97°  35',  h  =  27°  08'  22",  and  A  =  40°  51'  18",  to 
project  and  compute  the  triangle.    Between  Avhat  limits   must  the 


92 


SPHERICAL  TRIGONOMETRY. 


value  of  a  be  assigned  in  order  that  there  may  be  two  solutions  ?   Be- 
yond what  limiting  values  of  «  is  a  solution  impossible  ? 


Projection.     See  Fig.  65. 

Trigonometrical  Solution.— To  find  p,  sin  p 
=  sin  b  sin  A. 

log  sin  27°  08'  22"  =  9.659115 
+  log  sin  40°  51'  18"  =  9.815675 

Rejecting  10  =  9.474790  —  log  sin  p. 
B    .'.p  =  17°  21'  40".     [Reason  for  the  species.] 

TVe  now  observe  that  there  is  one  and  onli/  one 
solution,  since  the  arc  a  (97°  35')  cannot  lie  between 
CD  (17°  21'  40")  and  b  (27°  08'  22"),  but  can  lie  be- 


tween CD  aud  CA'  (180°  -  b  =  152°  51'  38"). 
To  find  AD,  cos  A  =  cot  5  tan  AD   or  tan  AD  = 


cos  A 
cot  6" 


log  cos  40°  51'  18"  =    9.878733 
-  log  cot  27°  08'  22"  =  10.290226 

Addinsr  10  =    9.5«8o07  =  log  tan  AD. 


AD  =  21°  11'  30".6. 
cos  5 


To  find  ansrle  ACD,  cos  b  =  cotA  cot  ACD,  or  cot  ACD  =  . 

=  '  cot  A 

log  cos  27°  08'  22"  =    9.949340 

-  log  cot  40°  51'  18"  =  10.0630575 

Adding  10 

[Reason  for  the  species.  ] 

smp 

sin  a 


9.8862825  =  log  cot  ACD.   .'.  ACD  =  52°  25'  01". 
To  find  B,  sin  p  =  sm  a  sin  B,  or  sui  B  = 


log  sin  p  (as  above)  =  9.474790 
-  log  sin  97°  35'  =  9.996185 

Adding  10  =  9.478605  =  log  sin  B.     . '. 
[Reason  tor  the  species.  ] 

cos  a 


B  =  17°  31'  09" 


To  find  DB,  cos  a  =  cos  p  cos  DB,  or  cos  DB  = 


cos  p 


log  cos  97°  35'         =  9.120469 
-  log  cos  17°  21'  40"  =  9  979750 

Adding  10  =  9.140719  =  log  cos  DB. 
[Reason  for  the  species.] 

AB  =  AD  +  DB  =  21°  11'  30".6  +  97°  56'  51".3  =  119°  08'  21".9. 


DB  =  97°  56'  51".3. 


sin  v 
To  find  DCB,  sin;?  =  tan  DB  cot  DCB,  or  cot  DCB  = — 


tan  DB" 
log  sin  p  (as  above)  =    9.474790 
—  log  tan  97°  56'  ol".3  =  10.855090 

Adding  10  =    8.619700  =  log  cot  DCB. 
[Reason  for  the  species.] 

ACB  ^  C  =  ACD  +  DCB  =  52°  25'  01"  +  92°  23'  7".7  =  144°  48'  8".7. 


DCB  =  92°  23'  7".7. 


OBLIQUE   ANGLED   TELA.NGLES   SOLVED   BY   NAPLER'S   RULE.        93 

Finally,  we  observe  that,  if  any  value  were  assigned  to  a  between  h  (27°  08'  22") 
and  CD  (17°  21'  40")  there  would  be  two  solutions ;  since  for  such  values  the  side 
a  could  lie  on  both  sides  of  CD.  But,  for  any  value  of  a  less  than  CD  (17°  21'  40"), 
there  would  be  no  solution  ;  since  CD  is  the  shortest  distance  from  C  to  the  arc 
ABA'.  Also,  for  any  value  of  a  greater  than  the  arc  CA'  (152°  51'  38"),  there 
would  be  no  solution,  as  such  an  arc  would  fall  between  CA'  and  CD'  (if  not 
>  CD'),  and  consequently  would  make  c  >  180". 

SCH. — Such  examples  as  this  and  the  preceding  can  be  more  expeditiously 
solved  by  using  p  in  each  equation  in  solving  the  triangles  ACD  and  DCB.  By 
this  means  and  using  {135)  to  determine  the  side  c,  the  solution  can  be  eflFected 
with  only  12  logarithms.     Thus  in  Ex.  2. 

1st.  To  find  i?,  Bin  p          =  sin  &  sin  Ai  require?   3  logarithms 

2d.   To  find  ACD,  cos  ACD  =  cot  &  tan 2?,    requires  3  " 

3d.    To  find  DCB,  cos  DCB  =  cot  «  tan;?,   requires  2  "       (log  tan p  being  known). 

4th.  To  find  B,  ^ini?          =  sin  a  sin  B,  requires  2  (log  sin  2?  being  known). 

5th.  To  find  c,      sin  A  :  sin  C  :  :  sin  a  :  sin  c,  requires  _2^        "       Gog's  of  sin  A  and  sin  a 

Total  12  logarithms  being  known.) 

•3.  Given  a  =  7G°  35'  3G",  1)  =  50°  10' 30",  and  c  =  40°  00'  10",  to 
project  and  solve  the  triangle. 
Projection. — See  Fij.  66. 

TRiGONO>rETRiCAL  SOLUTION. — Ist.  To  find  the 
segments  CD  and  DB,  we  have, 


tan  i{s  +  s'), 
or 
tan  ^a  :  tan  i{b  +  c)  :  :  tan  i(Z>  -  c)  :  tan  i{s  -  s'). 

f  \ 

Ja 

^<-^ 

/\ 

Computing  by  logarithms. 

"'Va 

\J   \\ 

a.c.  log  tan  ia          =  log  tan  38°  17' 48"  =    0.102561 

^  \ 

^^^^^^=^ 

+  log  tan  \{b  +  c)  =  log  tan  45°  05'  20"  =  10.001347 

+  log  tan  lib  -c)  =  log  tan    5^  05'  10"  =    8.949406 

Fig.  GG. 

Rejecting  10  =    9.053314  =  log  tan  i{s  -  s'). 

.-.  i(s_  5')  =  6°27'02". 

In  order  to  determine  whether  s  or  s'  is  the  greater,*  we  observe  the  signs 
of  the  proportion,  and  finding  tan  h{s  —  s')  positive,  know  that  s  >  s'. 

'   Hence,  s  =  \{s  +  s)  +  l{s  -  s')  =  38°  17'  48"  +  6°  27'  02"  =  44°  44'  50", 
and  s'  =  Ks  +  s')  -  K«  -  «')  =  38°  17'  48"  -  6°  27'  02"  =  31°  50'  46". 

The  angles   sought  are    now  readily  found  by  computing  the  two  right 
angled  triangles  ADC  and  ADB, 


*  Though  the  projection  generally  determines  such  facts  as  this,  the  species  of  parts,  and 
the  number  of  solutions  in  ambiguous  cases,  the  student  should  not  rely  upon  it,  but  determine 
each  such  fact  upon  purely  trigonometrical  considerations,  merely  using  the  projection  to  give 
clearness  to  the  conception,  and  as  a  rough  check  against  errors. 


94  SPHERICAL    TRIGONOMETRY. 

Or,  having  computed  C  from  the  triangle  ACD,  we  maj  find  A  and  B  more 
expeditiously  by  using  the  proportions, 

sin  c  :  sin  6  :  :  sin  C  :  sin  B, 
and  sin  c  :  sin  a  :  :  sin  C  :  sin  A. 

The  angles  are  C  =  34°  15'  03",  A  =  121°  36'  12",  and  B  =  42°  15'  13". 

4.  Given  A  =  128°  45',  C  =  30°  35',  and  a  =  68°  50',  to  solve  the 
triangle.    What  values  of  A  give  two  solutions  ?    What  none  ? 

c  =  37°  28'  20",  b  =  40°  09'  04",  and  B  =  32°  37'  58". 

5.  Given  A  =  129°  05'  28",  B  =  142°  12'  42",  and  C  =  105°  08'  10", 
to  solve  the  triangle. 

a  =  135°  49'  20",  b  =  146°  37'  15",  and  c  =  60°  04'  54". 

6.  Given  a  =  68°  46'  02",  b  =  37°  10',  and  C  =  43°  37'  38",  to 
project  and  solve  the  triangle. 

A  =  116°  22'  22",  B  ^  35°  29'  54",  and  c  =  45°  52'  34". 

7.  Given  a  =  40°  16',  b  =  47°  44',  and  A  =  52°  34',  to  project 
and  solve  the  triangle.  What  values  of  a  give  but  one  solution  ? 
What  none  ? 

There  are  hvo  triangles.— Ivl  the  1st,  c  =  53°  19'  20",  B  =  65°  16' 
35",  and  C  =  79°  52'  21".  In  the  2d,  c  =  14°  18'  22",  B  =  114°  43' 
25",  and  C  =  17°  39'  22". 

8.  Given  a  =  62°  38',  b  =  10°  13'  19",  and  C  =  150°  24'  12",  to 
project  and  solve  the  triangle. 

A  =  27°  31'  44",  B  =  5°  17'  58",  and   c  =  71°  37'  06". 

9.  Given  a  =  56°  40',  b  =  83°  13',  and  c  =  114°  30',  to- project 
and  solve  the  triangle. 

A  =  48°  31'  18",  B  =  62°  55'  44",  and  C  =  125°  18"  56". 

10.  Given  A  =  50°  12',  B  =  58°  08',  and  a  =  62°  42',  to  solve  the 
triangle.    What  values  of  A  give  but  one  solution  ?    What  none  ? 

There  are  Uoo  soMions.— 1st,  b  =  79°  12'  10",  c  =  119°  03'  26", 
and  C  =  130°  54'  28".  2d,  b  =  100°  47'  50",  c  =  152°  14'  18",  and 
C  =  156°  15' 06". 

11.  Given  A  =  36°  25',  B  =  42°  17'  10",  and  c  =  95°  10'  05",  to 
project  and  solve  the  triangle. 


OBLIQUE  ANGLED  TRIANGLES  SOLVED  BY  NAPIER's  RULES.   95 

12.  Given  a  =  124°  53',    b  =  31°  19',   and   c  =  171°  48'  22",  to 
solve  the  triangle. 

13.  Given  a  =  150°  17'  23",  h  =  43°  12',   and  c=  82° 50'  12,"  to 
solve  the  triangle. 

14.  Given  a  =  115°  20'  10",  b  =  57°  30'  06",  and  A  =  126°  37'  30", 
to  solve  the  triangle. 

15.  Given  A  =  109°  55'  42",  B  =  116°  38'  33",  and  C  =  120°  43' 
37",  to  solve  the  triangle. 

16.  Given  A  =  50°,  b  —  60°,  and  a  =  40°,  to  solve  the  triangle. 

17.  Given  a  ==  50°  45'  20",  b  =  69°  12'  40",  and  a  =  44°  22'  10", 
"^o  project  and  solve  the  triangle. 


There  are  fivo  solutions.— Ui,  B  =  57°  34'  51",  c  =  115°  57'  51", 
and  c  =  95°  18'  16".  2d,  B  =  122°  25'  09",  c  =  25°  44'  32",  and  c 
=  28°  45'  05". 

18.  Given  b  =  99°  40'  48",  c  =  100°  49'  30",  and  A  =  65°  33'  10", 
to  project  and  solve  the  triangle. 

a  =  64°  23'  15",  B  =  95°  38'  04",  and  c  =  97°  26'  29". 

19.  Given  A  =  48°  30',  B  =  125°  20',  and  c  =  62°  54',  to  solve  the 
triangle. 

a  =  56°  39'  30",  b  =  114°  29'  58",  and  c  =  83°  12'  06". 

20.  Given  c  =  54°  15'  03",  B  =  40°  18'  13",  and  a  =  70°  30'  30", 
to  solve  the  triangle. 

21.  Given  A  =  47°  54'  21",  c  =  61°  04'  56",  and  a  =  40°  31'  20", 
t@  project  and  solve  the  triangle. 

22.  Given  a  =  50°  10'  10",  b  =  69°  34'  35",  and  a  =  120°  30'  30", 
to  project  and  solve  the  triangle. 


96 


SPHEPJCAL  TilIGO^'OMETRY. 


SECTION   III. 

GOTHAL  formula:. 

[N'OTE.— This  section  is  designed  for  such  as 
make  matliematics  a  specialty.  The  preceding 
sections  are  thought  suflScieut  for  the  general 
student] 

14:3,  Prop. — In  a  Sj)herical  Tri- 
angle the  cosine  of  any  side  is  equal  to 
the  product  of  the  cosines  of  the  other  two 
sides,  plus  the  product  of  the  sines  of  those 
sides  into  the  cosine  of  their  included 
angle  j  that  is, 

(1)  cos  a  =  cos  h  cose  +  sin  b  sin  c  cos  A ; 

(2)  cos  I?  =  cos  a  cos  c  +  sin  a  sin  c  cos  B ; 

(3)  cos  c  =  eosa  cos  b  +  sin  a  sin  Z>  cos  C 


i 


Dem.— From  Fig.  67,  we  have, 
cos  a  =  cos  {c  —  x)  cos  p 

cos  (c  —  x)  cos  b     r  . 

=  ^: smce  cos 

cos  X  L 

cos  b  cos  c  cos  X  +  cos  b  sin  c  sin  x 

~~  cos  X 

=  COS  b  cos  c  +  cos  b  sin  c  tan  x       i 


_  cos  b  "I 
^  ~  cos  X  J 

[expanding. cos  (c  —  x)] 

] 


sm  X 

since  =  tan  x 

cos  a; 


cos  b  tan  x~ 


=  cos  b  cos  c  +  sin  J  sin  c  cos  A  |  since  cos  A  =  cot  6  tan  x=  ,       .. 

In  a  similar  manner  (2)  and  (3)  may  be  produced. 

144,  Cor.  l.—From  set  A,  hy  jjassiiig  to  the  polar  triangle,  loe 
have, 

(1)  cos  A  =  —  cos  B  cos  c  +  sin  B  sin  c  cos«  ;  \ 

(2)  cos  B  =  —  cos  A  cos  c  +  sin  A  sin  c  cos  &  ;  ^  B. 

(3)  cos  C  =  —  cos  A  cos  B  +  sin  A  sin  B  cos  c.    ) 


Dem.— Lettmg  a',  b\  d,  A',  B',  and  C  represent  the  parts  of  the  polar  triangle, 


GENERAL  FORMULAE. 


97 


we  have  a  =  180°  -  A',  6  =  180°  —  B',  c  =  180°  -  C,  A  =  180'  -  a\  B  = 
180°  —  b\  and  C  =  180°  —  e'.    Whence,  substituting  in  (1)  A,  we  have, 

cos  (180°  -  AO  =  cos  (180°  -  B')  cos  (180°  -  C) 

+  sin  (180°  -  B')  sin  (180°  —  C)  cos  (180°  -  «'), 
or,  cos  A'  =  —  cos  B'  cos  C  +  sin  B'  sin  C  cos  a', 

since  cos  (180°  -  A')  =  -  cos  A',  etc. ;  and  sin (180°  -  B)  =  sin  B',  etc. 

Finally,  dropping  the  accents,  since  the  results  are  general,  and  treating  (2) 
and  (8)  of  set  A  in  the  same  way,  we  have  set  B. 

14S,  Cor.  2. — From  A  a7icl  B  we  readily  find  tlie  angles  in  terms 
of  the  sides,  and  the  sides  in  terms  of  the  angles.     Thus,  from  A, 


,^ .  cos  a  —  cos  h  cos  c 

(1)  C0SA  = : — r—' ; 

^  sm  0  sm  c 

cos  h  —  cos  a  cos  c 


(2)  cosB 

(3)  cosC  = 


sm  a  sm  c 

cos  c  —  cos  a  cos  h 

sin  a  sin  ?> 


A'. 


rrom  B, 


,^  .  cos  A  +  COS  B  COS  c 

(1)  COS  a  = -. ■  — , 


(2)  COS  h  = 

(3)  cos  c  = 


sm  B  sm  C 
cos  B  -I-  cos  A  cos  c  . 

: : } 

sm  A  sm  C 

cos  C  +  cos  A  cos  B 

sin  A  sin  B 


>  B'. 


146,  JProp. — FormtdcB  A',  and  B',  adapted  to  logarithmic  coin- 
2mtation,  lecome. 


(1)  sin^A 

(2)  sin^B 


(3)  sin 


/sin 

(is 

—  b)  sin  (\s  - 

-'^■, 

t 

sin  b  sin  c 

A  /'''' 

iis 

—  a)  sin  (fsf  - 

-0). 

V 

sin  a  sin  c 

y 

i  /'^ 

(is 

—  a)  sin  {^s  - 

-b) 

sm  a  sm 
7 


M\ 


98 

And  (1)  sin 

(2)  sin  ^b 

(3)  sin  ^c 


SPHERICAL  TRIGONOMETRY. 


in  ^a  =  a/  - 


'        COS 

is  COS (is  - 

■A). 

sin  B  sin  c 

J 

COS 

is  cos (is  - 

B). 

sin  A  sin  c 

? 

COS  is  cos  (is  —  C  ) 
sin  A  sin  B 


B" 


Dem.— Subtracting  each  member  of  (1)  A'  from  1,  we  have, 

cos  a  —  cos  h  cos  c      cos  h  cos  c  +  sin  5  sin  c  —  cos  a  "" 


1  —  cos  A  =  1  — 
•.  2  sin"  ^A  = 


sin  b  sin  c 
cos  {b  —  c)  —  cos  a     . 


sin  5  sin  c 
,  since  1  —  cos  A  =  2  sin^  ^A  {62,  5), 


sin  b  sin  c 

and  cos  b  cos  c  +  sin  &  sin  c  =  cos  (6  —  c)  (55,  D). 

Now  letting  y  =  b  —  c,  and  x  =  «,  we  see  from  (59,  D')  that  cos  (6  —  c) 
—  cos  a  =  2  sin  ^(a  +  b  —  c)  sin  ^(a  +  c  —  b). 


Hence,  2  sin*  ^A 


2sini(a  +  b  —  c)  sini(ct  +  c  —  b) 
sin  6  sin  c  ' 


or, 


sin  ^  A 


_  .  /sin  i{a 


b  —  c)  sin  i(a  +  c  —  6) 


sin  6  sin  c 
Finally,  putting     s  =  a  +  b  +  c,  whence  K«  +  b  —c)  =  ^s  —  c, 


and 
we  have, 


x{a  +  c  —  b)  =  |s  —  6, 


„•„  lA        i  A^^  (2-"«  -  ^)  sin  (i«  -  c) 

sin  tA  =  4/  : — j-^ . 

r  sm  0  sin  c 


sin  b  sin  c 

In  like  manner,  (2)  and  (3)  of  set  A'  reduce  to  (2)  and  (3)  of  set  A". 
Again,  subtracting  each  member  of  (1)  set  B',  from  1,  we  have, 

cos  A  4-  cos  B  cos  C      sin  B  sin  C  —  cos  B  cos  C  —  cos  A 


1  —  cos  a  =  \ 


.'.  2sin'^  ia  = 


sin  B  sin  C  sin  B  sin  C 

—  cos  (B  +  C)  —  cos  A  _       cos  (B  +  C)  +  cos  A 


sin  B  sin  C  s         sin  B  sin  C 

2  cos  i(A  +  B  -i-  C)  cos  i(B  +  C^,A)  ^^^Jt, 


sin  B  sin  C 


GENERAL   FORMULA.  V 

Now  putting  S  =  A  +  B  +  C,  whence  i(B  +  C  —  A)  =  ^S  —  A,  we  have, 

.     ,  ,  /      cos  iS  cos  (iS  -  A ) 

sm  ia  =  i/ -. — fi—~^ 

r  sin  B  sm  C 

In  the  same  manner,  (2)  and  (3)  of  B"  are  deduced  from  (2)  and  (3)  of  b  . 
147.  Cor.  1. — Passing  to  the  polar  triangle,  A"  and  B"  become 


.^.  ,  ,   /cos  (is  —  B)  cos  (is  —  C). 

(1)  cos Uc  =  4/  — — : — —. — — -' 

^  ^         '^         1/  sm  B  sm  C 

(is  —  A)cos(-|S  -  C). 


1/  sm  A  sm  C 

_       /cos  (is  —  A)  cos  (is  —  b) 


sm  A  sm  B 


And 


(2)  cos  i^ 

(3)  cos  ic 

(1)  cos 

(2)  cosiB=|/- 

,„.  ,  ,  /sin  is  sin 

(3)  cos  ic  =  A/ ~. 

^  ^         ^         1/  sm  a 


^  B'" 


sm  i5  sm(i.s"  ~  ^0. 
sin  5  sin  c       ^ 


sin  i^  sin  (^s  —  b)  ^ 
sin  a  sin  6*       '  ' " 


sin  b 


14S»  ScH. — Formulae  A'"  and  B"  can  be  obtained  directly  from  A  and 
B,  in  a  manner  altogether  similar  to  that  in  which  A"  and  B"  were  deduced,  by 
adding  each  member  of  the  equations  in  sets  A'  and  B'to  1,  instead  of  subtract- 
ing, and  observing  that  1  +  cos.-c  =  2cos^  ^x. 

149,  Cor.  2. — Dividing  the  formitlm  of  set  k."  by  the  correspond- 
ing ones  of  set  A'";  and,  in  a  similar  manner,  those   o/ B'"    by 


those  ofB",  and  putting  ^^r.(y  -  a)  .m(y-J>),m(is  -  c) 


I: 


and  |/c"MiS-A)co.sas-B)coB(tS-c)  ^      ^^.^ 
'  .  —  cos  is 


(1)  tan 

(2)  taniB 


sin  {\s  —  b)  sin  (i-s^  —  c)  _  l: 

sin  is  sin  (is  —a)      ~  sin  {\s  —  a) ' 


sin  (is  —  a)  sin  (is  —  c)  _  k 

sin  is  sin  (is  —  b)       ~  sin  (is-Z>j 


(3)taniC=i/5HLii£zi4l4£lfc 
^  ^         ^        •!/         sm  is  sm  (is  — 


-b)  _  h 

c)       ~  sin  {\s  —  c)' 


100 


SPHERICAL  TRIGONO^IETRY. 


(1)  COtJrt 

(2)  cotib  =  A/' 


/cos (is 

y      -cos 


B)  COS  (jS  —  C)  _ 


^S  cos  (^s  —  A)         COS  (4s  —  A)  ' 

K 


cos  (^S  —  A)  COS  (Is  —  C) 


(3)  cotl-c 


/cos (is 
1/      -cos 


COS  is  COS  (is  —  B)         COS  (is  -  B) 

K 


B' 


a)  cos  (is  —  B) 


is  cos  (is  —  C)  cos  (is  —  C). 


ScH. — In  these  formulcB  k  is  the  tangent  of  the  arc  with  which  the  inscribed 
circle  is  described,  and  K   is  the  cotangent  of  the  arc  with  which  the  circum- 


FiG.  68. 


Fig.  69. 


scribed  circle  is  described.    Thus,  nsmg  the  common  notation,  we  have  In 
Fig.  68,    AD  =  AD'  =  \s  —  a,  and  angle  PAD  =  \^\  whence 

tan  PD 


sin  AD  =  cot  PAD  x  tan  PD 


tan  PAD' 


or  tan  ^A  = 


tan  PD 


k 


.,  [(l)Ai^].     .:k=  tanPD. 


sin  (^s  —  a)       sin  {{s  —  «)' 
From  Fig.  69,  we  have,  AD  =  ic,  and  angle  PAD  =  ^S  —  C. 

cos(iS  -C), 


Hence,       cos  (iS  —  C)  =  cot  AP  x  tanic,  or  tan  ^c 


cotAP 


or         cot  \c  = 


cot  AP 


cos  (is  -  C)     cos(iS  — C) 


,  [(3)Bi^].     .'.  K  =  cot  A  P. 


GAUSS'S  EQUATIONS. 

ISO,  I^rob, — To  deduce  Gauss's  Equations,  loliich  are 
sin  i(A  +  b)       cosi(rt  —  V) 


(1) 


cos  iC" 


cos  \c 


,  .   sini(A  —  B)  _  sini(rt  —  V)  ^ 
cosic       ~       siuic       ' 


(3) 


gauss's  equations.  101 

cos^(A  +  b)  _  cos-l{a  +  b) , 
smTj  C       ""       COS  ic       ' 

COS  |^(A  —  B)  _  sin  ii^i  +  i) 
sin  ^C        ""        sin  ^g 


Solution.— From  A,  page  25,  we  have, 

sin  (^A  +  iB),  or  sin  KA  +  B)  =  sin  ^A  cos  iB  +  cos  ^A  sin  ^B. 
Substituting  in  the  second  member  the  vahies  of  sin  ^A,  cos  iB,  cos  ^A,  and 
sin  iB  ,  from  A"  and  A'",  there  results, 


•    -,/«       r^x      sm{hs —  b)  . /smissmiis  —  c)      fiin  {U  —  a)     /sm  ^s  sm{is  —  C) 

smi(A  +  B)  = 7 4/  : : — 7 —   H -. i/ -. : — -. 

'  sm  c       r         sm  a  sm  b  sm  c        r        sm  a  sm  d 


_  sin  (jj?  -b)  +  sin  {js—a)     /sin  js  sin  {js  -  c)  ^, . 

sin  c  y         sin  6t  sin  6       '       '■  ^ ' 

cos  iC 


sm 

sin  (is  —  b)  +  sin  Qs  —  a) 


sm  c 

But  sin  (is  —  5)  +  sm  {is  —  a)z=  sin  {^a  +  |&  +  §c  —  &)  +  sin  {^a  +  ^b  +  ic  —  a) 

=  sin  [ic  +  i{a  —  b)]  +  sin  [^c  —  -iCa  —  b)] 
=  2  sin  ic  cos  ^^(a  —  6).     (59,  A'). 

Also,  sin  c  =  2  sin  |c  cos  ic. 

Substituting  these  values,  the  preceding  becomes, 

•     1/  A       ox       2  sin  |c  cos  Ua  —  b)         ,  _ 

sm  i(A  +  B)  = r-^ — ,    ^^    , ^  cos  iC  . 

2  sm  ic  cos  ic  ' 

sin  |(A  +  B)  _  cos  i{a  —  b) 
cos  iC        ~"       cos  ^c      •     w 

In  like  manner  starting  with 

sin  (|A  -  |B ),  or  sin  |(A  —  B)  =  sin  |A  cos  ^B  —  cos  -^A  sin  -JB, 

.,  ,^  sin  i(A  —  B)       sin  i{a  —  b)    ,^, 

there  results,  -—r?i — -  = ^^ ■     (2) 

cos  iO  sm  ic  ' 

Starting  with  cos  (iA  +  ^B),  or  cos  \{fK  +  B)  r=  cos  ^A  cos  ^B  —  sin  ^A  sin  ^B, 

there  results,  cos  «A  +  B)  ^  cosK«  ^) 

sm  -JC  cos  \o  ' 

Starting  with  cos  (^A  —  IB),  or  cos  \{k  —  B)  =  cos  M  cos  iB  +  sin  ^A  sm  ^B; 

*u  1*  cos  \{k  —  B)       sin  \{a  +  J)     ,,, 

there  results,  ^r— — '  = f^  r— -^    (4) 

sm  iC  sm  \c         ^  ' 


102  SPHERICAL  TRIGONOMETEY. 

NAPIER'S  ANALOGIES. 
lol.  I^roh, —  To  deduce  Napiefs  Analogies,  which  are 

(1) 


(2) 
(3) 


tan  |(a  +  b)  _  cos  ^{a  —  h)  ^ 
cot  -|-c        ~  cos  \{a  +  h) ' 

tan  |^(a  —  B)  _  sin  ^{a  —  h)  ^ 
cot  \C        ~~  sin  ^{a  +T) ' 

tan  ^{a  +  i)  _  cos  |(A  —  b). 
tan  |c        ~  cos  ^(A  +  B)' 

tan  ^(^—5)    _  sin^(A  —  B) 
tan  Jc       ""  sin4^(A  +  b)' 


Solution.— To  deduce  (1),  divide  the  1st  of  Gauss's  Equations  by  tlie  3d. 
To  deduce  (2),  divide  the  2d  of  Gauss's  by  the  4th.  To  deduce  (3\  divide  the  4th 
of  Gauss's  by  the  3d.    To  deduce  (4),  divide  the  2d  of  Gauss's  by  the  1st. 

152,  ScH. — In  using  these  fonnulse  the  species  must  be  carefully  attended 
to.  Thus  in  (1),  cot  iC  and  cos  \{a  —  h)  are  necessarily  +  ;  hence  tan  |(A  +  B) 
and  cos^(a  +  h)  are  of  the  same  sign  with  each  other.  In  (2),  cot  ^C  and 
sin  \{a  +  h)  are  both  +  ;  hence,  tan  |(A  —  B)  and  sin  \{a  —  b)  are  of  the  same 
sign  with  each  other.    And  similar  inspections  may  be  made  upon  (3)  and  (4). 


EXERCISES. 

153,  The  proposition  tliat  "  The  sines  of  the  angles  are  to  each 
other  as  the  sines  of  their  opposite  sides"  (133),  Napier's  Analogies 
(151),  and  formulae  A'%  B'"'  (140)  are  sufficient,  in  themselves,  to 
effect  the  solution  of  all  cases  of  oblique  spherical  triangles;  and 
for  practical  purposes  they  generally  require  less  labor  than  Xapier's 
Rules.  «We  give  a  few  solutions  and  refer  the  student  to  the  pre- 
ceding Exercises  for  further  practice. 

1.  Given  a  =  100°;  c  =  5°  and  h  =  10°,  to  solve  the  triangle. 
(Prob.  1,  Case  1st,  137.) 

1st.  To  find  A  and  B  we  have, 

cos  K«  +  *)  :  cos  K«  -  b)   :  :  cot  ^C  :  tan  i(A  +  B) ; 
and  sin  ^{a  +  b)  :  sin  ^{a  -  b)   :  :  cot  ^C  :  tau  i(A  -  B)     [15^  (1)  (2)] 


Napier's  analogies.  103 

Computing  by  logarithms,  we  have, 

ar.  CO.  log  cos  [K«  +  b)  =  55°]  =  0.241409 
+  log  cos  [i{a  -b)  =  45']  =  9.849485 
+  log  cot  [i  C  =  2°  30']      =  11.359907 

Rejecting  10  =  11.450801  =  log  tan  i{k  +   B). 

.-.  ^(A  +  B)  =  87"  58'  18". 
ar.  CO.  log  sin  [K«  +  b)  =  55°]  =    0.086635 
+  log  sin  [kia  -  b)  =  45°]  =    9.849485 
+  log  cot  [^C  =  2°  30']       =  11.359907 

Rejecting  10  =  11.290027  =  log  tan  ^{^  -  B). 

...  i(A  -  B)  =  87°  06'  16" 

The  signs  of  all  the  terms   being  + ,  |(A,  +  B)  and  ^(A  —  B)  are  both  less 
than  90°. 

KA  +  B)  +  i(A  -  B)  =  A  =  87°  58'  18"  +  87°  06'  16"  =  175°  04'  34" 
i(A  +  B)  -  i(A  -  B)  =  B  =  87°  58'  18"  -  87°  06'  16"  =      0°  52'  02". 

2cl.  To  find  c.    This  may  be  found  from  the  proportion, 
sin  A  :  sin  C  :  :  sin  a  :  sin  c, 
or  from  the  3d  or  4th  of  Napier's  Analogies.    We  use  the  last,  though  the  first 
is  equally  expeditious. 

sin  ^A  —  B)  :  sin  ^(A  +  B)  :  :  tan  i{a  —  b)  :  tan  ^c. 

ar.  CO.  log  sin  [^A  -  B)  =  87°  06'  16"]  =  0.000555 
+  log  sin  [KA  +  B)  =  87°  58'  18"]  =  9.099728 
+  log  tan  [\{a  -  b)   =  45°]  =  10.000000 

Rejecting  10  =  10.000283  =  log  tan  ^c. 

.-.  c  =  90°  02'  14". 

2.  Given  a  =  135°  05'  28".6,  c  =  50°  30'  08".6,  and  b  =  G9°  34' 
5G".2,  to  solve  the  triangle. 

1st.  To  find  a  and  c.    The  3d  and  4th  of  Napier's  Analogies  give, 
cos  KA  +  C)  :  cos  KA  —  C)  :  :  tan  lb  :  tan  Ka  +  c) ; 
and  sin  KA  +  C)  :  sin  KA  —  C)  :  :  tan  ^b  :  tan  l{a  —  c). 

Computing  by  logarithms,  we  have 

ar.  CO.  log  cos  [i(A  +  C)  =  92°  47'  48". 6]    =    1.3116286  *  ^ 

+  log  cos  [KA  -  C)  =  42°  17'  40"]      =    0.8690535 
+  log  tan  [id    =  34°  47'  28".l]  =    9.8418527 

Rejectmg  10  =  11.0225348  =  log  tan  Ua  +  c). 
.'.  ^{a  +  c)  =  95°  25'  25". 
i{a  +  c)>  90%  since  cos  KA  +  C)  is  -,  cos  KA  -  C)  is  +,  and  tan  ib  is  +, 
making  tan  K<^  +  c)  —. 

*  These  logarithms  are  taken  from  7-place  tables,  in  order  to  obtain  the  tenths  of  seconds 
accurately. 


104 


SPHERICAL  TRIGONOMETRY. 


ar.  CO.  log  sin  [KA  +  C)  =  92°  47'  48".6]  =  0.0005176 
+  log  sin  [KA  -  C)  =  42°  17'  40 "  ]  =  9.8279768 
+  log  tan  lib  =  34°  47'  28".  1]  =  9.8418527 


Rejecting  10  =  9.6703471  =  ] 


;  tan  ^{a  —  c). 
c)  =  25°  05'  05". 


lia  —  c)  <  90°,  since  the  signs  of  the  terms  are  all  + . 
K«  +  c)  +  i{a  -  c)=a  =  120°  30'  30",  and i{a  +  c)  —  i{a- c)  =  c  =  70° 20' 20". 

2d.  To  find  B.     Either  of  the  1st  two  of  Napier's  Analogies  will  give  B. 
Thus  (1)  becomes, 

cos  i(a  —  c)  :  cos  i{a  +  c)  :  :  tan  ^(A  +  C)  :  cot  ^B  ; 
and  (2)  sin  i{a  —  c)  :  sin  ^a  +  c)  :  :  tan  i(A  —  C)  :  cot  ^B- 

But  as  i{a  +  c)  is  so  near  90°,  it  will  be  better  to  use  the  second  of  these 
than  the  first.    Or  we  may  with  equal  accuracy  use, 

sin  c  :  sin  5  :  :  sin  C   :  sin  B. 

ar.  CO.  log  sm  (c  =  70°  20'  20"  )  =  0.0260878 
+  log  sm  (6  =  69°  34'  56". 2)  =  9.9718202 
+  log  sin (C  =  50°  30'  08".6)  =  9.8874210 

Rejectmg  10  =  9.8853290=  log  sm  B.    .-.  B  =  50°  10'  10". 

3.  Given   a  =  50°  45'  20",  b  =  69°  12'  40",   and  A  =  44°  22'  10", 
to  solve  the  triangle. 


1st.  To  find  B. 


sin  a  ;  sin  5  ; ;  sin  A  :  sin  B. 


ar.  CO.  log  sin  {a  =  50°  45'  20'')  =  0.1110044 

+  log  sin  {b  =  69°  12'  40 ")  =  9.9707626 

+  log  sm(A=  44°  22'  10")  =  9.8446525 

Rejecting  10  =  9.9264195  =  log  sin  B.  .-.  B  =  57°  34'  51".4, 
and  122°  25'  08".6.  There  are  two  solutions,  since  a  is  intermediate  in  value 
between  _?;  and  both  b  and  180°  —  b* 


*  The  determination  of  the  species  of  B,  or  what  is 
the  same  thing,  the  number  of  solutions,  can  usually  be 
effected  by  a  simple  inspection  without  any  computa- 
tion whatever.  Thus,  sin  j9  =  sin  5  sin  A.  the  loga- 
rithms of  which  are  given  above,  as  ie  log  sin  a.  Now, 
as  both  a  and  p  are  <  90°,  and  log  sin  p  <  log  sin  a, 
p  <:a.  But  a  <  6,  and  also  Icsa  than  180°  —  b.  All 
this  can  be  seen  at  a  glance. 


Napier's  analogies.  *  105 

To  find  C  and  c  of  the  larger  triangle  in  which  B  =  57°  34'  51".4 
Napier's  1st  gives 

ar.  CO.  log  cos  [i(6  -  a)  =    9°  13'  40"]     =    0.0056570 
+  log  cos  [i(6  +  a)  =  59°  59']  =    9.6991887 

+  log  tan  [KB  +  A)  =  50°  58'  30".7]  =  10.0912464 

Rejecting  10  =    9.7960921  =  log  cot  iC 

.-.  C=  115°  57' 50".  7. 

Napier's  3d  gives 

ar.  CO.  log  cos  [|(B  -  A)  =  6°  36'  20".7]  =  0.0028928 
+  log  COS  [i(B  +  A)  =  50°  58'  30".7]  =  9.7991039 
+  log  tan  [i(6  +  a)   =  59°  59'J  =  10.2382689 

Rejecting  10     =  10.0403656  =  log  tan  ^. 

.-.  c  =  95°  18'  16".4, 

3d.  To  find  C  and  c  of  the  smaller  triangle  in  which  B  =  122°  25'  08".6. 
Using  the  same  formulm  as  before. 

ar.  CO.  log  cos  [\{h  -  a)  -    9°  13'  40"]    =    0.0056570 
+  log  cos  \\{b  +  a)  =  59°  59']  =    9.6991887 

+  log  tan  [i(B  +  A)  =  83°  23'  39".3]  =  10.9362703 

Rejecting  10  =  10.6411160  =  log  cot  iC 

.-.  C  =  25°  44'  31".6. 

ar.  CO.  log  cos  [KB  -  A)  =  39°  01'  29".3]  =  0.1096506 
+  log  cos  [KB  +  A)  =  83°  23'  39".3]  =  9.0008369 
+  log  tan  [K^»  +  a)    =  59°  59']  =   10.2382689 

Rejecting    10        =      9.4087564  =  log  tan  ^c. 

.-.  c  =  28°  45'  05".2. 

154.  ScH. — "When  Napier's  Analogies  are  used  for  solving  Pro&.27i(Z  (159),  tlie 
most  expeditious  and  elegant  method  of  resolving  the  ambiguity,  is  by  means 
of  the  analogies  themselves.  Thus,  in  the  above  example,  after  having  found 
that  B  =  57°  34'  51". 4,  or  122°  25'  08".  6,  or  both,  a  simple  inspection  of  the  anal- 
ogy next  used  will  determine  the  number  of  solutions. 

Napier's  1st  may  be  written 

••  ,^       cos  K^  +  «) ,     ,,„ 

cot  iC  =: ^ r-^  tan  KB  +  A). 

cos  i{b  —  a)         ^  ' 

Now  iC  <  90°,  hence  cot  ^C  is  + .  If,  therefore,  neither  of  the  values  of 
B  renders  cot  iC  —,  there  are  two  solutions.  If  one  value  renders  cot^C  +, 
and  the  other  — ,  there  is  one  solution  and  it  corresponds  to  the  value  of  B 
which  makes  cot  ^C  +.  If  both  values  of  B  render  cot  ^C  — ,  there  is  no  solu- 
tion. In  the  last  example,  we  see  that  cos  [K^  +  «)  =  59°  59],  and  cos  [^(6  —  «) 
=  9°  13'  40"]  are  both  + .  Also  tan  [KB  +  A)  =  50°  58'  30". 7,  or  83°  23'  39".3, 
or  both]  is  +  for  both  values  of  B.     Therefore  there  are  two  solutions. 


106  '  SPHEKICAL  TRIGONOMETRY. 

4.  Given  A  =  95°  16',  B  =  80°  42'  10",  and  a  =  57°  38',  to   solve 
the  triangle. 

1st.  To  find  &,  sin  A   :  sin  B   :  :  sin  a  :  sin  b. 

ar.  CO.  log  sin  (A  =  95'  IG)  =  0.001837 
+  log  sitt  (B  =  80°  43'  10")  =  9.994257 
+  log  sin  (a  =  57°  38)         =  9.926671 

Rejecting    10  =  9^922765  ==  log  sin  h. 

:  b  =  56°  49'  57",  or  123°  10'  03",  or  both. 

2d.  To  find  c,  tan  ic  =  ^^i|^-±-^  tan  i(a  +  b).    Now  for  b  =  56°  49'  57", 

COS  ^v**  —   ^/ 

tan  |c  is  +  ;  but  for  b  =  123'  10'  03 "  tan  ^c  is  —  ;  hence  there  is  but  one  solu- 
tion, and  that  corresponds  to  the  smaller  value  of  b. 

ai-.  CO.  log  cos  [KA  -  B)=  7°  16'  55"]  =  0.003517 
+  log  cos  [i(A  +  B)  =  87°  59'  05"]  =  8.546124 
+  log  tan  [K«  +  b)-  57°  13'  58"]  =  10.191.352 

Rejecting    10  =    8.740993  =  log  tan  U. 

.-.  c  r=  6'°  18' 19". 

3d.  To  find  C,  we  may  use  (1)  or  (2)  of  Napiei-'s  Analogies,  or 
sin  «  :  sin  c  :  :  sin  A  :  sin  C, 
the  last  of  whicn  is  the  most  expeditious. 

ar.  CO.  log  sin  {a  =  57°  38)  =  0.073329 
+  log  sin  (c  =  6'  18'  19")  =  9.040705 
+  log  sin  (A  =  95°  16')         =  9.998163 


Rejecting    10  =  9.112197  =  log  sin  C  .-.  C  =  7°  26'  22' 
This  value  is  taken  for  C  instead  of  its  supplement,  since  C  is  opposite  the 
smallest  side  c. 

5.  Given  a  =  70°  14'  20",  h  =  49°  24'  10",  and  c  =  38°  46'  10";  to 
solve  the  triangle. 


COMPUTATIOK. 

a  =  70°  14'  20" 
b  =  49°  24  10" 
c  =  38'  46'  10" 
5  =  158°  24  40" 


is  =    79°  12'  20"    ar.  co.  log  sin  =       0.007753 
^s-  a  =     8°  58'  00"  "     "    =       9.192734 

is-  b  =    29°  48'  10"  "     "    =       9.696370 

U-  c  =    40°  26'  10"  "     "    =       9.811977 

2)18.708834 


.-.  log  k  =       9.354417 


Napier's  analogies.  107 

log  tan  iA  =  \ogk-  log  sin  {is  -  a)  +  10=  10.161683.  .-.  A  =  110"  51  16". 
log  tan  iB  =  log  A;  -  log  sin  (is  -  5)  +  10  =  9.658047.  .-.  B  =  48*  56'  04". 
log  tan  iC  =\ogk  -  log  sin  {is  —  c)  +  10  =    9.542440.     .-.  C  =    38°  26'  48". 

6.  Given  a  =  109°  55'  42",  B  =  116°  38'  33",  and  c  =  120°43'  37", 
to  solve  the  triangle. 

COMPUTATION. 

A  =  109°  55'  42" 
B  =  116°  38'  33" 
C  =  120°  43'  dl" 
S  =  347°  ir  52" 

iS  =  173°  38'  56"  ar.  co.  log  cos  =       0.002683 
iS  -  A  =    63°  43'  14"  "      "   =       9.646158 

iS  -  B  =   57°  00'  23"  "      "    =       9.730035 

iS  -  C  =   52°  55'  19"  "      "    -       9.780247 

2)19.165123 
.-.  log  K  =        9.582561 

log  cot  ia  =  log  K  -  log  cos  (iS  ~  A)  +  10  =  9.936403.  .-.  a  -  98°  21'  38". 
log  cot  \h  =  log  K  -  log  cos  (iS  -  B)  +  10  r=  9.846526.  .-.  b  =  109°  50'  20". 
log  cot  ie  =  log  K  -  log  cos  (^S  -  C)  +  10  =  9.802314.    .\c  =  115°  13'  28". 

ScH.  1.— The  student  can  use  the  exercises  in  the  preceding  section  to  famil- 
iarize the  metliods  here  given.  In  doing  so,  it  will  be  well  for  him  to  seek  the 
most  expeditious  soUitious.    He  will  find  that 

Examjiles  iinder  Prob.  1  require  11  logarithms  by  Napier's  Analogies  and 
{135),  and  12  logarithms  by  Napier's  Rules  and  {135). 

Examples  under  Pjrob.  2,  when  there  is  but  one  solution,  require  10  loga- 
rithms by  Napier's  Analogies  and  {135),  and  12  logarithms  by  Napier's  Rules 
and  {135).  When  there  are  two  solutions,  15  logarithms  are  required  by 
Napier's  Analogies  and  {135),  and  only  14  by  Napier's  Rules  alone,  or  by  these 
rules  and  {135). 

Examples  under  Prob.  3  require  but  7  logarithms  by  the  method  given  in  this 
section  and  13  by  the  previous  method. 

ScH.  2. — In  cases  in  which  the  angles  or  sides  are  near  the  limits  0°,  90°,  or 
J80°,  so  that  the  functions  used  in  the  particular  solution  change  very  rapidly 
in  proportion  to  the  arc,  it  is  often  possible  to  select  one  among  the  several 
methods  which  will  give  more  accurate  results  than  the  others.  There  are 
also  other  methods  which  are  better  adapted  to  such  cases  than  those  here 
given.  For  these,  as  well  as  for  much  other  interesting  matter,  and  especially 
for  the  discussion  of  the  General  Spherical  Triangle,  American  students  have  an 
excellent  resource  in  the  treatise  of  Professor  Chauvenet  of  Washington  Univer- 
sity, St.  Louis. 


108  SPHEEICAL  TKIGONOMETRY. 

SECTION  IV, 

AREA  OF  SPHERICAL  TRIA^SGLES. 

155,  JProh, — Having  the  angles  of  a  S2}lierical  triangle  given,  to 
find  the  area. 

Solution. — [The  solution  is  given  in  Pabt  II.  {613),  and  we  simply  re- 
produce the  result  in  order  to  give  completeness  to  this  section.]  The  area  is 
equal  to  the  ratio  of  the  spherical  excess  to  90°,  or  ^tt,  into  the  trirectangular  tri- 
angle. That  is,  letting  the  sum  of  the  angles  be  S°,  the  area  K,  and  the  radius  of 
the  sphere  1,  whence  the  area  of  the  trkectangular  triangle  is  ^tt,  we  have 

K  =  ^-^  X  ^7t  =  S-n. 
In  the  latter  expression  S  is  the  sum  of  the  angles  in  terms  of  the  radius,  i.  e.j 

CO  CO 

^  =  5r:29578'  ^'  ^PP^o^i^^tely,  S  =  .-^  {9). 


EXERCISES. 

1.  "What  is  the  area  of  a  spherical  triangle  whose  angles  are  100°, 
58°,  and  62°,  on  a  sphere  whose  diameter  is  6  feet  ? 

Solution.     K  =  S  —  ;r  =  -^  —  3.14159  =  .698,  the  area  of  a  similar  ti'i- 

angle  on  a  sphere  whose  radius  is  1.  Hence,  the  area  of  the  required  triangle 
is  .698  X  3*  =  6.282.  [The  method  given  in  Pakt  II.  (613)  is  more  expedi- 
tious, but  it  is  our  purpose  to  illustrate  the  form  here  given.] 

2.  What  is  the  area  of  a  si3herical  triangle  whose  angles  are  170°, 
135°, and  115°,  on  a  sphere  whose  radius  is  10  feet? 

Ans.  418.875  square  feet. 

3.  What  is  the  area  of  a  spherical  triangle  whose  angles  are  150°, 
110°,  and  G0°,  on  a  sphere  whose  radius  is  3  feet  ? 


ISO.  J^rob. — Having  the  sides  of  a  sjyherical  triangle  given,  to 
find  the  area. 

Solution. — The  angles  may  be  found  by  {14:8)^  and  then  the  area  by  {15S). 


AREA  OF  SPHERICAL  TRIANGLES.  109 

But  a  more  direct  method  is  to  find  the  spherical  excess  by  means  of  LhuiU 
Iter's  formula,  which  we  will  now  produce. 

^K   =  :l(A     +    B  +   C  -  TT) 

Whence  tan  ^K  =  tan  ^[A  +  B  +  C  -  :nr]  =  tan  [KA  +  B)  -  \(7t  -  C)] 

_  sin  i(A  +  B)  -  sin  \{Tt  —  C) 

~  cos  KA~+  B)  +  cos  \{Ti  -  C)  '        (7,  page  3t\ 

_  sin  i(A  +  B)  —  cos  ^C 
~  cos  i(A  +  B)  +  sin  iC 

=  [coaK^-6)-cos-k]cosiC_  ^^        ^^^  ^^^  3^^^ 

[cos  \{(i  +  6)  +  cos  \c\  sin  iC 


_  cos  \{a  —  b)  —  cos  |c     /     sin  js  sin  {U  —  c)  (146   Id"^) 

~~  cos  ^a  +  b)  +  cos  ic  r   sin  (|s  --  a)  sin  Us  —  b)  ' 


sin  i(a  +  c  -  &)  sin  i{b  +  c  —  a)     /     sin  -^^  sin  (-^g  —  c)  „   p,  ^, 

cos  i(a  +  b  +  c)  cos  i(a  +  6  —  c)    r    sin  (il-s  —  a)  sin  (U  —  b)  '      ' 


sin=^  -Kis  —  6)  sin^  His  —  a)  sin  |8  sin  (is  —  c)  ,        _  a  .    \ 

cos'  :i«  cos'^  i(i.s  —  c)  sin  (^s  —  a)  sin  (^s  —  5)  '       "" 


sin*^  ii^s  —  d)  sin^  i{U  —  a)  sin  is  cos  ^s  sin  jd^  —  c)  cos  -Ki^  —  <^)    ,^^v 
cos''  is  cos=^  i(is— c)  sin  i(is— a)  cos  i(is— «)  sin  i{is—b)  cos  i(i«— 6) 


.-.  Tan  iK  =*  .y/tan  is  tan  i(|s  -  a)  tan  i(is  -  b)  tan  i(i«  -  c).  (A) 

Having  found  K,  the  spherical  excess,  or  what  is  the  same  thing,  the  area  of 
a  similar  triangle  on  a  sphere  whose  radius  is  1,  we  have  but  to  multiply  K  by 
the  square  of  the  radius  in  any  given  case. 


EXERCISES. 

1.  Given  a  =  98°,   b  =  110°,  and  c  =  115°,  to  find  the  area  of  a 
spherical  triangle,  on  a  sphere  whose  radius  is  4000  miles. 

COMPUTATION". 

log  tan  {\s  =  80"  45')  =  10.788185 

+  log  tan  [{\s-ia)  =  31°  45']  =  9.791563 
+  log  tan  [{is-U)  =  25°  45]  =  9.683356 
+  log  tan  [{\s-ic)  z=  23°  15]  =    9.633098 

2)39.896203 
Rejecting  10  =    9.948101  =  log  tan  ^K  .    .-.  K  =166°  20'  20" 

*  The  +  is  always  taken ;  otherwise,  iK  being  >  90°,  K  would  be  i>  360'  which  is  impoa- 
eible.    (Part  III.,  256). 


110  SPHERICAL   TRIGONOMETRY. 

Whence  area  =  i?^^— '  .  irt  (4000)'  =  ^^  .  i^  (4000)»  =  46,450,440, 
nearly. 

2.  Given  a  =  70°  14'  20",  5  =  49°  24'  10",  and  c  =  38°  46'  10",  to 
find  the  area  of  a  spherical  triangle  on  a  sphere  whose  diameter  is  8 
feet.  -^fis.  5.1,  nearly. 


lo7.  I*roh. — Having  fico  sides  and  their  included  angle  given 
in  a  spherical  tria^igle,  to  find  the  area. 

Solution.— Compute  the  other  two  angles  by  Napier's  Analogies,  and  find 

rtM     ^         ,        ^  ,  ^      cot  ^  cot  ^  +  cos  C    . 
the  area  from  the  angles.     [The  formula  cot  ^K  = = r^^ gives 

the  spherical  excess  in  terms  of  two  sides  and  their  included  angle  ;  but  it  is  of 
no  practical  value  for  finding  the  area,  as  it  is  not  adapted  to  logarithmic  compu- 
tation. For  the  manner  of  producing  it  and  several  other  forms  for  K,  see 
Todhunter's  Spherical  Trigonometiy,  {103)]. 


PRACTICAL    APPLICATIONS. 

[Note. — The  three  following  problems  are  given  merely  to  indicate  to  the 
student  some  departments  of  investigation  in  which  Spherical  Trigonometry  is 
of  essential  service.  The  two  sciences  to  which  this  branch  of  Pure  Mathe- 
matics is  indispensable,  are  Geodesy,  or  the  mathematical  measurement  of  the 
earth,  and  Astronomy.] 

^roh,  l.—To  find  the  shortest  distance  on  the  earth^s  surface  be- 
tween two  2^oi7its  whose  latitudes  and  longi- 
tudes are  knoiun. 

Sug's. — The  shortest  distance  on  the  surface  be- 
tween two  points  is  the  arc  of  a  great  circle  joining 
the  points.  Hence,  the  Problem  is  :  Given  two  sides 
(the  co-latitudes)  and  the  included  angle  (the  differ- 
ence in  longitude),  to  find  the  third  side. 


Fig.  71.  Ex.  1.  Berlin  is  situated   in    Lat.  52°  31' 

13"  N.,  Lon.  13°  23'  52"  R,  and  Alexandria,  Egypt,  in  Lat.  31°  13' 
X.,  Lon.  29°  55'  E.  Wliat  is  the  shortest  distance  in  miles  on  the 
earth's  surface  between  them,  the  earth  being  considered  a  sphere 
whose  radius  is  3962  miles  ? 

Ans.  1G91.96  miles. 


PRACTICAL  APPLICATIONS. 


Ill 


Ex.  2.  A  ship  starts  from  Valparaiso,  Chili,  Lat.33°  02'  S.,  Lon.  71° 
43'  W.,  and  sails  on  the  arc  of  a  great,  circle  in  a  northwesterly  direc- 
tion 3840  miles,  when  her  longitude  is  found  to  be  120°  W.  What 
is  her  latitude  ?  Ajis.  52'  48"  S. 

Ex.  3.  A  ship  starts  from  Eio  Janeiro,  Brazil,  Lat.  22°  54'  S.,  Lon. 
42°  45'  W.,  and  sails  in  a  northeasterly  direction  on  the  arc  of  a  great 
circle  624.44  miles,  when  her  latitude  is  found  to  be  50°  N.  What  is 
her  longitude  ?  Ans.  2°  01'  14"  W. 


JProb,  2. — To  find  the  time  of  day  frcmi  the  altitude  of  the  sun, 

Sug's.— Let  NESQ  represent  the  projection  of  the  concave  of  the  heavens 
upon  the  plane  of  the  meridian  of  observation. 
The  equator  of  this  concave  sphere  is  simply  the 
intersection  of  the  plane  of  the  earth's  equator  with 
this  imaginary  concave  sphere,  and  its  axis  is  the 
prolongation  of  the  earth's  axis,  the  poles  being 
the  points  N  and  S  where  the  axis  pierces  the 
imaginary  concave.  EQ  is  the  projection  of  this 
celestial  equator,  NS  the  axis  or  the  projection  of 
a  great  circle  perpendicular  to  the  meridian  of  the 
observer  (NESQ)  and  to  the  equator,  HO  the  projec- 
tion of  the  horizon,  and  ZZ'  the  projection  of  the 
prime  vertical  (that  is,  a  great  circle  of  the  heavens 
passing  through  the  zenith  of  the  observer  and  the  east  and  west  points  in  his 
horizon). 

Now  let  S'  be  the  place  of  the  sun  at  the  time  of  observation.  RS',  the  sun's 
declination,  is  known  from  the  almanac ;  LS',  the  sun's  altitude,  is  measured 
with  the  sextant  (or  other  instrument) ;  and  EZ  is  the  latitude  of  the  observer. 
Hence,  in  the  spherical  triangle  ZNS'  we  know  the  three  sides,  viz.,  NS'  =  the 
co-declination  of  the  sun,  ZS'  =  the  co-altitude  of  the  sun,  and  ZN  =  the  co- 
latitude  of  the  observer.  We  may  therefore  compute  the  angle  ZNS',  which 
reduced  to  time  gives  the  time  before  or  after  noon  as  the  case  may  be. 

Ex.  1.  On  April  21st  the  master  of  a  ship  at  sea  in  latitude  SO"* 
OG'  20"  N.,  observed  the  altitude  of  the  sun's  centre  at  a  certain  time 
in  the  forenoon  and  found  it  to  be  30°  10'  40",  and  looking  in  the 
almanac  found  the  sun's  declination  at  that  time  to  be  12°  03'  10" 
N.    What  was  the  time  of  day  ? 


COMPUTATION. 


90°  -  30°  10'  40"  = 
90°  -  12°  03'  10"  = 
90°  —  39°  06'  20"  = 


59°  40'  20" 
77°  56'  oO" 
50°  53'  40" 
2)188°  39'  50" 
94°  19'  55 " 


a.  c.  log  sin  94°  10'  55"  ~ 

a.  c.  log  sin  34°  30'  35"  = 

log  sin  16°  23'  05"  = 

log  sin  43°  26'  15"  = 


0.001242 
0.246765 
9.450381 
9.837312 
2)19.535:0') 
log  tan  30°  22'  03".3  =   9.707650 


112  SPHERICAL  TRIGONOMETRY. 

Therefore  i  the  hour  angle  NNS'  =  30'  2Z'  03  '.3,  and  the  hour  angle  is  60° 
44'  07".  This  reduced  to  time  at  4  minutes  to  a  degree,  gives  4  h.  2  m.  56  s.  be- 
fore noon,  or  7  h.  57  m.  4  s.  a.  m. 

Ex.  2.  In  latitude  40°  21'  X.,  when  the  declination  of  the  sun  is 
3°  20'  S.,  and  its  altitude  36°  12',  what  is  the  time  of  day  ? 

Ans.  9  h.  42  m.  40  s.  A.  M. 

Ex.  3.  In  latitude  21°  02'  S.,  when  the  sun's  declination  was  18° 
32'  X.,  and  the  altitude  in  the  afternoon  40°  08',  what  was  the  time 
of  day?  Ans.  2  h.  3  m.  57"  P.  m. 


JProb.  S,—To  fijul  the  time  of  sunrising  and  sunsetting  at  any 
given  place  on  a  given  day. 

Sug's. — The  projection  being  the  same  as  before,  let  M'RS'M  represent  the  ap- 
parent diurnal  path  of  the  sun.    Since  S'M  is  described  in  6  hours,  the  time  taken 

to  describe  RS'  is  the  time  before  6  o'clock,  at 
which  the  sun  rises,  i.  e.,  passes  the  horizon  HO. 
But  the  time  requisite  to  describe  RS',  is  the  same 
part  of  24  hours  (360'  angular  measure)  that  the 
angle  CNL  (=  arc  CL)  is  of  360".  Hence,  the  arc 
CL,  in  time,  is  the  time  before  6  o'clock  at  which 
the  sun  rises.  In  a  similar  manner,  C/,  in  time, 
is  seen  to  be  the  time  after  6  o'clock  when  the  sun 
is  south  of  the  equator.  The  solution  of  the  prob- 
lem, therefore,  consists  in  finding  CL.  Now,  in 
Fig.  73.  the  triangle  RLC,  right  angled  at  L,  LR  =  the 

sun's  declination  at  the  time,  and  angle  RCL  =  ECH  =  the  co-latitude  of  the 
place.*    From  these  data  CL  is  readily  found. 

Ex.1.  Eequired  the  time  of  sunrise  at  latitude  42°  33'  N.,  when 
the  sun's  declination  is  13°  28'  N. 

COMPUTATIOK". 

cot  47°  27'         =  9.962813 

tan  13°  28'         =  9.379239 

sin  12'  41'  52"  =  9.342052 
(12°  41'  52")  X  4  gives  the  time  before  6  o'clock  as  50'  47".    .'.  The  sun  rises  at 
5  h.  09  m.  13  s. 

*  This  may  be  seen  thus :  Suppose  a  person  to  start  from  the  equator  at  E  and  travel 
north.  When  he  is  at  E.  the  south  point  of  hia  horizon  (H)  is  at  S  ;  and  for  every  degree  he 
goes  north,  the  south  pole  (S)  sinks  a  degree  below  his  horizon.  Hence,  HCS  =  ^is  latitude, 
and  ECH  =  co-latitude. 


PRACTICAL  APPLICATIONS.  113 

Ex.  2.  Required  the  time  of  sunrise  at  latitude  57°  02'  5-4"  X., 
when  the  sun's  declination  is  23°  28'  JST. 

Sun  rises  at  3  h.  11  m.  49  s. 

Ex.  3.  How  long  is  the  sun  above  the  horizon  in  latitude  58°  12' 
N.,  when  the  sun's  dechnation  is  18°  41'  S.,  that  is  about  Januaiy 
25th  ?  Ans.  7  h.  35  m.  36  s. 

Ex.  4.  What  is  the  length  of  the  longest  day  at  Ann  Arbor,  Mich., 
Lat.  42°  16'  48".3,  the  sun's  greatest  declination  being  23°  27'  ? 

Ans.  15  h.  05  m.  50  s. 

[Note. — In  such  problems  as  the  foregoing,  several  small  corrections  have  to 
be  made  in  order  to  entire  accuracy,  such,  for  example,  as  that  for  refraction  in 
taking  the  altitude,  and  for  the  time  required  for  the  sun's  disk  to  pass  the  hori 
zon.    But  they  would  be  out  of  place  here.  ] 


THE   EIs-D. 


INTRODUCTION 


TO    THE 


TABLE   OF    LOGARITHMS 


[Note. — If  the  student  understands  the  nature  and  use  of  logarithms  so  as  to 
be  able  to  use  the  common  tables,  it  will  not  be  necessary  that  he  should  read 
this  introduction.  Otherwise  a  careful  study  of  it  will  be  needful  before  reading 
Section  4  of  the  Plane  Trigonometry.] 

i.  A  Lofjavith/in  is  the  exponent  by  w^liich  a  fixed  number  is 
to  be  affected  in  order  to  produce  any  required  number.  The  fixed 
number  is  called  the  Base  of  the  System. 

III.  Let  the  Base  be  3 :  then  tho  logarithm  of  9  is  2 ;  of  27,  3 ;  of  81,  4 ;  ol 
19623,  9  ;  for  3"  =  9 ;  3=*  =  27 ;  3*  =  81 ;   and  3'  =  19083.     Again,  if  64  is  the 

base,  the  logarithm  of  8  is  -J-,  or  .5,  siuce  64",  or  64'  =  8 ;  i.  e.,  |,  or  .5  is  the 
exponent  by  which  64,  the  base,  is  to  be  affected  in  order  to  produce  the  num« 

ber  8.  So  also,  64  being  the  base,  ^,  or  .333  + ,  is  the  logarithm  of  4,  since  64  ,  or 
g^.33  3  +  _  4.  ^  g^  1^  Qj,   333  ^^  ig  tij(j  exponent  by  which  64,  the  base,  is  to  be 

f 
affected  in  order  to  produce  the  number  4.     Once  more,  since  64  ,  or  64''^^  -^  = 

16,  !,  or  .666  +,is  the  logarithm  of  16,  if  the  base  is  64.  Finally,  64~"  or 
64~"'  =  I,  or  .125 ;  hence  —  |,  or  —  .5,  is  the  logarithm  of  \,  or  .125,  when  the 
base  is  64.  In  like  manuer,  with  the  same  base,  —  i,  or  —  .333  +,is  the  loga- 
rithm of  i,  or  .25. 

EXAMPLES. 

1.  If  2  is  the  base  what  is  the  logarithm  of  4  ?  of  8  ?  of  32  ?  of 
128?  of  1024? 

Solution.  7  is  the  logarithm  of  128,  if  2  is  the  base,  since  7  is  the  exponent 
by  which  2  is  to  be  affected  in  order  to  produce  the  number  128. 

2.  If  5  is  the  base,  what  is  the  logarithm  of  625?  of  15625?  of 
125?  of  25? 


a  INTRODUCTION  TO  THE   TABLE   OF  LOGARITHMS. 

3.  If  10  is  the  base,  what  is  the  logarithm  of  100  ?  of  1000  ?  of 
10,000?  of  10,000,000  ? 

4.  If  2  is  the  base,  what  is  the  logarithm  of  J,  or  .25  ?  of  -J,  or  .125  ? 
of  ^,  or  .03125  ?  Ans.  to  the  last,  —  5. 

5.  If  8  is  the  base,  of  what  number  is  |,  or  MQ  +  the  logarithm? 
of  what  number  is  |-,  or  1.333  +,  the  logarithm  ?  of  what  number  is 
2  the  logarithm?  of  what  number  is  2\,  or  2.333"  +  ?  of  what  num- 
ber 3|,  or  3.666  +  ?  Ans.  to  the  last,  2048. 

ScH.  Since  any  number  with  0  for  its  exponent  is  1,  the  logarithm  of  1  is  0 
in  all  systems.  Thus  10*  =  1,  whence  0  is  the  logarithm  of  1,  in  a  system  in 
which  the  base  is  10. 


2.  A  System  of  LoffCirWuns  is  a  scheme  by  which  all  num- 
bers can  be  represented,  either  exactly  or  approximately,  by  expo- 
nents by  which  a  fixed  number  (the  base)  can  be  affected. 

3.  There  are  Ttvo  Systems  of  Logarithms  in  common  use,  called, 
resi^ctively,  the  Briggean  or  Common  System,  and  the  Kajyierian 
or  Hyperlolic  System.  The  base  of  the  former  is  10,  and  of  the 
latter  2.71828  -f.  In  speaking  of  logarithms  of  numbers,  the  com- 
mon logarithm  is  always  signified,  if  no  specification  is  made. 

4.  The  logarithms  of  all  numbers,  except  the  exact  powers  of  the 
base,  indicate  a  power  of  a  root,  and  are  consequently  fractional  and 
usually  only  approximations.  It  is  customary  to  write  them  in  the 
form  of  decimal  fractions.  The  integral  part  is  called  the  Char- 
acteristic,  and  the  fractional  part  the  3Iantissa.  The  charac- 
teristic can  always  be  told  by  a  simple  inspection  of  the  number 
itself;  hence  only  the  mantissa  is  commonly  given  in  the  table. 


5»  JProp. — TJie  characteristic  of  the  common  logarithm  of  any 
mimber  greater  than  icnity,  is  one  less  than  the  number  of  integral 
figures  in  the  given  number. 

III.  The  logarithm  of  4685  is  more  than  3,  because  10'  =  1000,  and  less  than 
4,  because  10*  =  10,000;  hence  it  is  3  +  a  fraction.  The  same  method  may  be 
pursued  to  determine  the  characteristic  of  the  logarithm  of  any  other  number 
greater  than  unity,  and  the  truth  of  the  proposition  be  observed.  Thus  the 
logarithm  of  2o645.827  is  4,  since  the  number  lies  between  the  4th  and  5th 
powers  of  the  base,  10. 


6*.  I^rojy* — Tlie  mantissa  of  a  decimal  fraction^  or  of  a  mixed 
number,  is  the  same  as  the  mantissa  of  the  number  considered  as 
integral. 


INTRODUCTION   TO   THE   TABLE   OF  LOGARITHMS.  3 

Dem.     Suppose  it  is  known  log  2845672  =  6.454185.      This  means   thai 

10'*'*"*  _  2845672.    Dividing  by  10  successively,  we  have 

1Q5.454185  ^  284567.2,  or  log  284567.2*         =  5.454185, 

10* •*"»"»  =    28456.72,  or  log    28456.72         =  4.454185, 

^Qs.-,54i86  ^      2845.672,        or  log      2845.672        =  3.454185, 
10'-*"*"  =        284.5672,      or  log        284.5672      =  2.454185, 
10>-45«ie.  ^     >    28.45672,    or  log         28.45672    =  1.4.54185, 
•    io°-""«^  =  2.845672,  or  log  2.845672  =  0.454185. 

Now  if  we  continue  the  operation  of  division,  only  writing  0.454185  —  1, 

1.454185,  meaning  by  this  that  the  characteristic  is  negative  and  the  mantissa 

positive,  and  the  subtraction  not  performed,  we  have 

lQ-464 186  ^  0.2845672,      or  log  0.2845672      =1.454185, 
lQ»:464ie6  ^  0.02845672,    or  log  0.02845672    =  2^454185, 
lQ^464i85  ^  0.002845672,  or  log  0.002845672  =  8.454185, 
etc.,  etc.    Q.  E.  D. 

7.  Cor.  T7ie  characteristic  of  a  numler  consistiiig  e^itirely  of  a 
decimal  fraction,  is  negative,  and  one  more  than  the  numler  of  O's 
i7nmecliately  following  the  decimal  point,  as  appears  from  the  last 
demonstration,  or  from  the  fact  that  1"*  =  -jig-  =  .1 ;  10~'  =  ^hr  = 
.01 ;  10-"  =  T^  =  .001 ;  etc.,  etc. 

8,  One  of  the  most  important  uses  of  logarithms  is  to  facilitate 
the  multiplication,  division,  involution,  and  extraction  of  roots  of 
large  numbers.  These  processes  are  performed  upon  the  following 
principles : 

9»  l?rop»  1. — The  sum  of  the  logarithms  of  two  numbers  is  thb 
logarithm  of  their  product. 

Dem.  Let  a  be  the  base  of  the  system.  '  Let  m  and  n  be  any  two  numbers 
whose  logarithms  are  x  and  y  respectively.  Then  by  definition  «■=  =  m^  and 
ay  =  n.  Multiplying  these  equations  together  we  have  a^-^y  =  mn.  Whence 
a;  +  y  is  the  logarithm  ofmn.    q.  e.  d. 

10.  JProj)*  2. — The  logarithm  of  the  quotient  of  two  numbers  is 
the  logarithm  of  the  dividend  minus  the  logarithm  of  the  divisor. 

Dem.  Let  a  be  the  base  of  the  sj'^stem,  and  m  and  n  any  two  numbers  whose 
logarithms  are,  respectively,  x  and  y.     Then  by  definition  we  have  rt^  =  m,  and 

a"  =  n.    Dividing,  we  have  a^-v  =  _.    Whence  x-y\^  the  logarithm  of  ^-. 

Q.  E.  D. 


*  This  is  the  common  abbreviation  indicating   the    logarithm  of  a  number,  and  should  bo 
read  "logarithm  of  284567.2,"  not  ''log  284567.2,"  which  ie  grossly  inelegant. 


i  INTKODUCTION  TO  THE  TABLE   OF  LOGAEITH^diS. 

11,  I^i'Oj),  3. — TJie  logarithm  of  a  jjoiuer  of  a  number  is  the 
logarithm  of  the  number  7mdtiplied  by  the  index  of  the  jJOicer. 

Dem.  Let  a  be  tlie  base,  and  x  the  logarithm  of  m.  Then  cF  =  m\  and  raising 
both  to  any  power,  as  the  2th,  we  have  a^*  =  wr.  Whence  xz  is  the  logarithm 
of  the  2th  power  of  m.    q.  e.  d, 

12,  I^rop,  4, — The  logarithm  of  any  root  of  a  number  is  the 
logarithm  of  the  number  divided  by  the  number  exjjressing  the  degree 
of  the  root. 

Dem.    Let  a  be  the  base,  and  x  the  logarithm  of  m.    Then  a^  =  m.    Extract* 

-      2  —  2"  2/— 

ing  the  2th  root  we  have  a""—  \/m.     Whence  -  is  the  logarithm  ofVm.    Q.  e.  d. 


TABLE  OF  LOGARITIDIS. 

IS.  In  order  to  apply  the  above  principles  practically,  we  need 
what  is  called  a  Table  of  Logarithms.  That  is,  a  table  from  which 
we  can  readily  obtain  the  logarithm  of  any  number,  or  the  number 
corresponding  to  any  logarithm.  Such  a  table  is  given  on  pages  11 
to  28.  For  methods  of  computing  it,  the  student  is  referred  to 
algebra.  Its  nature  and  manner  of  use  will  be  learned  from  the 
two  following  problems : 


14.  Prob,—  To  fiid  the  logarithm  of  a  number  from  the  table. 

Solution.  The  logarithm  of  any  number  hehceen  1  and  100  is  seen  directly 
from  the  table  on  page  11.  The  column  marked  N  contains  the  numbers,  and  the 
adjacent  column  to  the  right  gives  the  logarithm  of  the  corresponding  number 
\o  G  places  of  fractious.     Thus,  log  7  =  0.845098 ;  log  33  =  1.518514. 

The  mantissa  of  the  loganthm  of  any  number  exjJi'cssed  with  three  figures  is 
found  in  the  column  headed  0,  on  some  one  of  the  pages  from  12  to  26  inclusive. 
The  given  number  being  found  in  the  column  marked  N,  the  mantissa  of  its 
logarithm  stands  opposite.  Tlie  characteristic  can  be  determined  by  (5),  (6),  (7). 
When  but  four  figures  are  found  opposite  the  number  in  the  0  column,  the  two 
left-hand  figures  of  the  mantissa  are  the  same  as  in  the  next  mantissa  above, 
which  has  six.    Thus,  log  443  =  2.646404. 

To  find  the  logarithm  of  a  number  consisting  of  four  figures.  Let  it  be  required 
to  find  the  logarithm  of  293G.  Looking  for  293  (the  first  three  figures)  in  tlie 
column  of  numbers,  and  then  passing  to  the  right  until  reaching  the  column 
headed  6,  the  fourth  figure,  we  find  7756,  to  wliich  prefixing  the  figures  4f5, 
\7iiich  belong  to  all  the  logarithms  following  them  till  some  others  are  indicated, 
We  have  for  the  mantissa  of  the  logarithm  of  2936,  .467756.     But,  as  3  is  the 


INTRODUCTION  TO  THE  TABLE  OF  LOGARITHMS.         5 

logarithm  of  1000,  and  4  of  10,000,  log  3930  is  3  and  this  decimal,  or  log  2936  = 
3.467756. 

Tojind  the  logarithm  of  a  number  consisting  of  more  than  four  figures.  Let  it 
be  required  to  find  the  logarithm  of  2815672.  Finding  the  decimal  part  of 
logarithm  of  the  first  four  figures  2845,  as  before,  we  find  it  to  be  .454082.  Now 
the  logarithm  of  2846  is  153  (millionths,  really)  more  than  that  of  2845,  as  found 
in  the  right-hand  column,  marked  D.  Hence,  assuming  that  if  an  increase  of  the 
number  by  1000  makes  an  increase  in  its  logarithm  of  153,  an  increase  of  672  in 
the  number  will  make  an  increase  in  the  logarithm  of  tWtj,  or  .672  of  153,  or 
103,  omitting  lower  orders,  and  adding  this  to  .454082,  we  have  .454185  as  the 
mantissa  of  log  2845672.  The  integral  part  is  6,  since  2845672  lies  between  the 
6th  and  7th  powers  of  10.    Hence,  log  2845672  =  6.454185.    q.  e.  d. 

ScH.  1.  If  in  seeking  the  logarithm  of  any  number,  any  of  the  heavy  dots 
noticed  in  the  table  are  passed,  their  places  are  to  be  filled  with  O's,  and  the  first 
two  figures  of  the  decimal  of  the  logarithm  taken  from  the  0  column  in  the  line 
below.  Thus,  log  3166  is  3.500511.  This  arrangement  of  the  table  is  merely 
a  convenient  method  of  saving  space. 

ScH.  2.  The  column  marked  D  is  called  the  column  of  Tabular  Differences  ; 
and  any  number  in  it  is  the  difference  between  the  mantissas  found  in  columns 
4  and  5,  which  is  usually  the  same  as  between  any  two  consecutive  logarithms 
in  the  same  horizontal  line.  The  assumption  made  in  using  this  difference,  viz., 
that  the  logarithms  increase  in  the  same  ratio  as  the  numbers,  is  only  approxi- 
mately true,  but  still  is  accurate  enough  for  ordinary  use. 

EXERCISES. 

1.  Find  the  logarithm  of  2200.  .......  Logarithm,  3.342423. 

2.  Find  the  logarithm  of  24.3G Logarithm,  1.38G677. 

3.  Find  the  logarithm  of  2.698 Logarithm.,  0.431042. 

4.  Find  the  logarithm  of  3585.9 Logarithm,  3.554598. 

5.  Find  the  logarithm  of  42.6634 Logarithm,  1.630050.-1 

6.  Find  the  logarithm  of  331.957 Logarithm,  2.521082.^.^ 

7.  Find  the  logarithm  of  2519.38 Logarithm,  3.401294.  -^ 

8.  Find  the  logarithm  of  .538329 Logarithm,  1.731047. 

,  9.  Find  the  logarithm  of  .087346 Logarithm-,  2.941243. 

10.  Find  the  logarithm  of  .007389 Logarithm,  3.86858G. 

15,  ScH.  3.  It  will  be  observed  that  the  tabular  difference  varies  rapidly  at 
the  beginning  of  the  table;  hence,  for  numbers  between  10000  and  11000  it  is 
better  to  use  the  last  two  pages  of  the  table. 


IG.  JProh   2. — To  find    a  number    corresponding  to  a   given 
hviarithm. 


6  INTEODUCTION   TO   THE   TABLE   OF  LOGARITHMS. 

Solution.  Let  it  be  required  to  find  the  number  corresponding  to  the 
iofrarithm  5.5152G4  Looking  in  the  table  for  the  ?iext  less  mantissa,  we  find 
.515211,  the  number  corresponding  to  which  is  3275  (no  account  now  being 
taken  as  to  whether  it  is  integral,  fractional,  or  mixed  ;  as  in  any  case  the  figm'es 
will  be  the  same).  Now  from  the  tabular  difference,  in  column  D,  we  find  that 
an  increase  of  133  (million ths,  really)  upon  this  logarithm  (.515211),  would  make 
an  increase  of  1  in  the  number,  making  it  3276.  But  the  given  logarithm  is 
only  53  greater  than  this,  hence  it  is  assumed  (th  ^ugh  only  approximately 
correct)  that  the  increase  of  the  number  is  ■f'h  of  1,  or  53  -r- 133  =  .3984  +  . 
This  added  (the  figures  annexed)  to  3275,  gives  32753984  +  .  The  characteristic, 
being  5,  indicates  that  the  number  lies  between  the  5th  and  6th  powei^s  of  10, 
and  hence  has  6  integral  places.    .'.  5.515264  =  log.  327539.84  +  .     Q.  e.  d. 

EXERCISES. 

1.  Find  the  number  wliose  logarithm  is  1.240050. 

Number,      17.38. 

2.  Find  the  number  whose  logarithm  is  2.431203. 

Number,      269.9. 

3.  Find  the  number  whose  logarithm  is  3.503780. 

Number,  3189.91. 

4.  Find  the  number  whose  logarithm  is  0.138934. 

Number,      1.377. 

5.  Find  the  number  whose  logarithm  is  1.368730. 

_  Number,  .233738. 

6.  Find  the  number  whose  logarithm  is  2.448375. 

^Number,  .028078. 

7.  Find  the  number  whose  logarithm  is  3.538630. 

Number,  .003456. 

8.  Find  the  number  whose  logarithm  is    .  843970. 

_  Num.ber,  6.98184. 

9.  Find  the  number  whose  logarithm  is  1.867372. 

iVl^mJer,  .736837. 

10.  Find  the  number  whose  logarithm  is   .003985. 

_  Number,  1.00921. 

11.  Find  the  number  whose  logarithm  is  ¥.  723460. 

Number,  .005290. 


APPLICATIOS 

1.  Find,  by  means  of  logarithms,  the  product  of  57.98  by  18. 

SoLUTiojr.    As  the  logarithm  of  the  product  equals  the  sum  of  the  logarithms 
of  the  factors  \9\  we  find  the  logarithms  of  57.98,  and  18  from  the  table,  and 


INTRODUCTION   TO   THE   TABLE   OF  LOGARITHMS.  7 

adding  Ihern  together,  find  the  number  corresponding  to  the  sum.    The  latter 
number  is  the  product  required.     Thus, 

log  57.98  =  1.763278 
log  18      =  1.22o273 

3.018551  =  log  1043.64. 

2.  Multiply  23.14  by  5.062. 

3.  Multiply  0.00563  by  17. 

4.  Multiply  397.65  by  43.78. 

5.  Multiply  0.3.854  by  0.0576. 

6.  Find  the  continued  product  of  3.902,  597.16, 

and  0.0314728. 

7.  Multiply  832403  by  30243. 

8.  Multiply  9703407  by  90807. 

9.  Multiply  3.47  by  9.83. 
10.  Multiply  12.763  by  10.976. 

[Note.   The  examples  in   division  below  will  offer  additional  exercise,  if 
necessary.] 


Prod. 

117.1347. 

Prod. 

0.09571. 

Prod. 

17409.117. 

Prod. 

0.022199. 

M6, 

Prod. 

73.3354. 

Prod. 

25174363929.* 

Prod. 

881137279449. 

Prod. 

34.1101. 

Prod. 

140.086688.* 

1.  Divide  24163  by  4567. 

Solution.  Since  the  logaiithm  of  the  quotient  equals  the  logarithm  of  the 
dividend  minus  the  logarithm  of  the  divisor,  we  have  the  following  operation  t 

log  24163  =  4.383151 
log    4567  -  3.659631 

0.723520  =  log  5.29078, 
which  number  is  the  quotient. 

2.  Divide  56.4  by  0.00015. 

Operation,  log       56.4  =  1^751279 
log  0.00015  =  4176091 
Difference  of  log's  =  5.575188     .*.  The  quotient  is  376000. 

SuG.  Observe  that  only  the  cJiaracteristic  of  the  logaritlun  is  negative,  and 
that  in  subtracting  we  are  to  regard  the  nature  of  the  logarithmic  numbers  a« 
positive  or  negative. 

3.  Divide  461.02876  by  21.4. 

4.  Divide  25.49052  by  2.46. 

5.  Divide  17610.8248  by  37.6. 

6.  Divide  .00144  by  1.2. 

7.  Divide  .0000025  by  .005. 


A71S. 

21.5434. 

Alls. 

10.362. 

Ans. 

468.373. 

A71S. 

.0012. 

Ans. 

.0005. 

8  •         I^'TRODUCTION   TO   THE   TABLE   OF   LOGARITHMS. 

8.  Divide  43.2  by  .24.  Ans.  ISO 

0.  Divide  59.74514  by  1.36.  Ans.  43.93025. 

10.  Divide  .0001728  by  2.4.  A)is.    .000072. 

[Note.  The  examples  in  multiplication  given  above  will  afford  additional 
exercise,  if  necessarj',]    . 

17.  ^cu.— Arithmetical  Comjyleinefit.—The  arithmetical  complement 
of  a  number  is  simply  the  remainder  after  subtracting  the  number  from  some 
particular  fixed  number.  Thus,  the  a.  c.  of  5  with  reference  to  9  is  4 ;  of  3. 6 ;  of 
7,  2;  etc.  The  a.  c.  of  7  with  reference  to  10  is  3  ;  of  4,  6 ;  of  2,  8  ;  etc.  When 
required  to  subtract  one  number  from  another,  we  may,  if  we  choose,  add  its  a.  c. 
and  then  subtract  the  number  with  reference  to  which  the  a.  c.  is  taken.  This 
process  will  give  the  true  difference.  Thus,  if  we  are  to  subtract  6  from  9,  we 
may  add  to  9  what  6  lacks  of  being  10  (10  —  6  =  4,  the  a.  c.  of  6  with  reference 
to  10)  and  then  subtract  10.  9  —  6  =  9  +  4  —  10.  A  few  such  questions  as  the 
following  will  render  this  simple  process  familiar.  What  number  must  I  add  to 
57G,  in  order  that  I  may  subtract  100  from  the  sum,  and  get  the  same  remainder 
as  if  I  had  subtracted  58  in  the  first  instance  ?  Again,  if  I  wish  to  take  87  from 
160,  what  must  I  add  to  the  latter,  in  order  that  I  may  subtract  40  from  the 
result,  and  get  the  difference  sought  ? 

This  principle  is  sometimes  used  in  computing  by  means  of  logarithms.  It  is 
especially  convenient  when  there  are  several  multipliers  and  several  divisors 
involved  in  the  same  computation.  An  example  or  two  will  make  the  process 
familiar.  The  complements  of  logarithms  are  usually  taken  with  reference  to 
10.    If  the  logarithm  exceeds  10,  20  may  be  used,  etc. 

18.  Eeqnired  the  result  of  the  following  combinations :  346  X  27.8 
-^  1156  X  3426  ~  2.004  X  27  -^  11.05. 

Operation.  log    846  =  2.539076 

log  27.8  =  1.444045 
a.  c.  log  1156  =  6.937042 
log  3426  =  8.534787 
a.  c.  log  2.004  =  9.698102 
log      27  =  1.431864 
a.  c.  log  11.05  =  8.956638 
84.541054 
Rejecting  30.000000  as  three  complements    are 

used.  4.541054  is  the  logaritlim  of  the  re- 

quired result.     .-.   As  4.541054  =  log  34757.92,  the  latter  is  the  result  sought 

[Note.  The  preceding  examples  can  be  used  to  familiarize  this  principle  if 
thought  desirable.] 

Suo.  The  a.  c.  of  2.468216  is  11.531784,  since  2  is  negative.  An  a.  c.  can  bo 
written  directly  from  the  table  with  nearly  the  same  ease  as  the  logarithm 
itself,  by  writing  from,  left  to  right,  and  taking  each  figure  from  9,  except  the 


INTRODUCTION  TO   THE   TABLE   OF   LOGAKITHMS.  9 

riglit-liand  one,  which  is  to  be  taken  from  10.    Thus,  if  the  characteristic  is  3, 
we  write  6  •  the  next  figure  being  2,  write  7  ;  for  4,  write  5,  etc. 


1.  What  is  the  cube  of  32  ? 

Solution.  Since  the  logarithm  of  the  cube  of  a  number  is  three  times  the 
logarithm  of  the  number  itself  {IT),  we  have 

»  log  (32)'  =  3  log  32  =  4.515450  =  log  32767.9T, 

which  number  is  the  cube  of  32,  as  accurately  as  the  process  gives  it.    (32)'  by 
multiplication  =  32768. 

2.  What  is  the  cube  root  of  7896.34? 

Sua  Log  (7896.34)^  =  ^  log  7896.34  -  1.299142  =  log  19.913.  .'.  (7896.34)^ 
=  19.913.    (Seei;^.) 

3.  What  is  the  20th  power  of  1.06  ?  Ans,    3.2071. 

4.  What  is  the  5th  poAver  of  2.846  ? 

5.  What  is  the  5th  root  of  31152784.1  ?  Ans.  31.52+. 

6.  What  is  the  cube  root  of  30  ?  Ans.  3.107  + . 

7.  What  is  the  cube  root  of  .03  ? 

SuG.  Log  .03  ="2.477121.  Now  to  divide  this  by  3,  we  have  to  bear  in  mind 
that  the  characteristic  alone  is  negative;  i.  e.,  2.477121  =  — 2 +_.477121,  or — 
1.522879.  This  divided  by  3  gives  -  .507626,  or  0  -  .507626  =  1.492374.  But 
a  more  convenient  metliod  of  effecting  this  division  is  to  write  for  the  —  2, 
-3  +_1,  whence  we  have  for  "2.477121,-3  +  1.477121,  which  divided  by  3 
gives  1.492374,  nearly. 

8.  Divide  3^261453  by  2,  by  4,  by  5.  Last  quotient,  1.4522906. 

9.  What  is  the  4th  root  of  .00000081  ?  Ans,       .03. 
10.  What  is  the  7th  root  of  0.005846?                              Ans.   .4707. 


1.  If  28.035  :  3.2781  :  :  3114.27  :  x,  what  logarithmic  operations 
will  fiud  X  ? 

'SuG.  The  logarithm  of  the  product  of  the  means  is  the  sum  of  their  loga- 
rithms ;  and  the  logarithm  of  the  quotient  of  this  product  divided  by  the  first 
extreme,  is  the  logarithm  of  said  product  minus  the  logarithm  of  the  other 
extreme.  .-.  log  x  =  log  3.2781  +  log  311427  -  log  28.035  =  0.515622  +  3.493356 
-  1.447700  =  2.561278.    Hence,  x  =  364.1478  + .  ^ 

2.  Given  72.34  :  2.519  :  :  357.48  :  x,  to  find  x,  by  logarithms. 

X  =  12.448. 

3.  Given  6853  :  489  :  :  38750  :  .r,  to  find  x,  by  logarithms. 

a:  =  2765. 


10  INTRODUCTION  TO   THE   TABLE  OF  LOGAKITHMS. 

SuG.  The  most  elegant  way  to  solve  such  propositions  by  logarithms  is  to 
take  the  sum  of  the  logarithms  of  the  means  and  the  a.  c.  of  the  given  extreme 
and  reject  10,    The  result  is  log  x. 

4.  Given  497  :  1891  :  :  37G  :  x,  to  find  re,  using  the  a.  c.  log. 

Operation.  log    1891  =  3.37G693 

log     37G  =  2.575188 
a.  c.  log     497  =  7.303644 
Sum,  less  10  =  3.155524  =  log  1430.62.    .'.  x  =,1430.63. 

[Note.   Solve  the  preceding  in  a  similar  manner,  by  using  a.  c.  log.] 


Let  the  student  give  the  reasons  for  the  following : 

1.  Given  (|)2  ^  {^Y  =  x,  we  have 

2  log   2  =  0.602060 

'i  log  16  =  0.903090 
a.  c.  2  log  3  =  9.045758 
a.  c.  Uog  5  =  9.475773 
Sum,  less  20  =  0.026681.     .-.  x  =  1.0033+  . 

2.  Given  V\  :  x  :  :  (3^)2  :    ^/¥,  to  find  x. 

Log  x  =  2  log  3  +  i  log   2  +  §  log  6  +  a.  c.  2  log  10  +  a.  c.  i  log  5  -  20  = 
.274039.    .-.  cc  =  0.1879  +  . 

3.  Given  \/Tl5  X  V^'^  :  (0.0051)'  :i  x  : j^^v 

Log  a;  =  i  log  115  +  h  log  .016  +  ^  log  .32  +  a.  c.  f  log  1146  +  a.  c  2  log 
0051  -  20  =  2.729701.    .-.  x  =  536.66  +. 


TABLE    I 


CONTAINING 


lOGAIlITHMS  OF  NUMBEES 


Feom  1  TO  11,000. 


N. 

Log. 

N. 

Log. 

N. 

Log. 

'  N. 

Log. 

1 

0-000000 

26 

1-414973 

51 

1-707570 

j  76 

1.880814 

2 

0'3oio3o 

27 

i-43i364 

52 
63 

1-716003 

77 

1-886491 

3 

0-477I2I 

28 

1-447158 

1-724276 

1  78 

1-892095 

4 

o.6o2o6o 

1  29 

1-462398 

54 

1-732394 

79 

1-897627 

5 

0-698970 

30 

I  •47-7121 

55 

1-740363 

80 

1-908090 

6 

0-778151 
0-845098 

31 

1-491362 

56 

1-748188 

1 
81 

1-908485 

7 

32 

1 -505150 

57 

1-755875 

82 

1-913814 

8 

0-903090 

33 

i-5i85i4 

58 

1-763428 

83 

1-919078 

9 

0-954243 

34 

i-53i479 

59 

1-770852 

84 

1-924279 

10 

I • 000000 

35 

1-544068 

60 

1.778151 

85 

1-929419 

11 

1-041393 

36 

1 -556303 

61 

1-785330 

86 

1-934498 

12 

1-079181 

37 

1-568202 

62 

1-792892 

87 

1 -989519 

13 

1-113943 

38 

1-579784 

.  63 

I -799341 

88 

1-944483 

14 

1-146128 

39 

1-591065 

64 

1-806180 

89 

1-949890 

15 

1-176091 

40 

1-602060 

65 

1-812913 

90 

1.954243 

16 

1-204120 

:  41 

1-612784 

66 

I -819544 

91 

1.959041 

17 

1-230449 

1  42 

1-623249 

67 

1-826075 

92 

1-968788 

18 

1-205273 

43 

1-633468 

68 

1-832509 

1-3 

1-968483 

19 

1-278754 

44 

1-643453 

69 

1-838849 

94 

1-978128 

20 

1 -301030 

45 

1-653213 

70 

1-845098 

95 

1-977724 

21 

1-322219 

46 

1-662758 

71 

I-85I258 

96 

I -982271 

22 

1-342423 

47 

1-672098 

72 

1-857333 

97 

1-986772 

23 

1-361728 

48 

I -681 241 

73 

1-863323 

98 

1-991226 

24 

1-3802II 

49 

1-690196 

74 

1-869282 

,  99 

1-995635 

25 

1-397940 

50 

1 -698970 

75 

1-875061 

100 

2  -  000000 

Remark.— In  the  following  Table,  the  first  two  figures,  in  the  first  column  of 
Logarithms,  are  to  be  prefixed  to  each  of  the  numbers,  in  the  same  horizontal 
line,  in  the  next  nine  columns;  but  when  a  point  (•)  occurs,  a  0  is  to  be  put 
in  its  place,  and  the  two  initial  figures  are  to  be  taken  from  the  next  line  below. 


12 


LOGARITHMS   OF   NUMBEKS. 


N. 

1  ' 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

100 
101 
102 
103 
104 
105 
106 
107 
108 
109 

1   4321 

86oo 

012837 

7033 

i  021 iSo 

;   53o6 

1   o384 

03J424 

7426 

0434 
4751 

f.^ 

7451 
i6o3 

57.5 
0789 
3826 
7825 

0868 
5i8i 
0451 
368o 
7868 
2016 
6125 
•195 
4227 
8223 

i3oi 
5609 
9876 
4100 
8284 
2428 
6533 
•600 
4628 
8620 

1734 
6o38 
•3oo 
4521 
8700 
2841 
6942 
1004 
5029 
9017 

2166 
6466 
•724 
4940 
9116 

3252 

7350 
1408 

5430 
9414 

2598 
6894 
1147 
5360 
9532 
3664 

7737 
1812 
5830 
9811 

^029 
7321 
1D70 
5779 

9947 
4073 
8164 
2216 
623o 
•207 

3461 
7748 
1993 

'.It] 
4486 
8571 
2619 
6629 
•602 

3891 
8174 
24i5 
6616 

4896 
8978 

3021 

7028 

•998 

432 
428 
424 

f^6 

412 
408 
404 
400 
396 

110 
111 
112 
113 
lU 
115 
116 
117 
113 
119 

041393 
5323 
9218 

053078 
6905 

060698 
4438 
8186 

071882 
5547 

1787 
5714 
9606 
3463 
7286 
1075 
4832 
8557 

2250 

5912 

2182 
oio5 

3«46 
7666 
1452 
5206 
8928 
2617 
6276 

2576 
6495 
•38o 
423o 
8046 
1829 
558o 

%n 

6640 

•766 
46i3 
8426 
2206 
5953 
9668 
3352 
7004 

3362 

88o5 

2582 

6326 
••38 
3718 
7368 

3755 
7664 
i538 
5378 
9185 
2958 
6699 
•407 
4o85 
773i 

4148 
8o53 
1924 
5760 
9563 
3333 
7071 
•776 
445 1 
8094 

4540 
8442 
2309 
6142 
9942 
3709 
7443 
1 145 
4816 
8457 

4932 

883o 
2694 
6524 

•320 

4o83 
7bi5 
i5i4 
5i82 
8819 

393 

IP. 

382 

372 
363 

120 
121 
122 
123 

124 
125 
126 
127 
123 
129 

079 1 81 

082785 
636o 
9905 

093422 
6910 

100371 
3804 
7210 

1 1 0390 

9543 
3i44 
6716 
•258 
3772 
7257 
0715 
4146 
7549 
0926 

4122 
7604 
1059 

4487 
7888 
1263 

•266 
3861 
7426 
•963 
4471 
7951 
i4o3 
4828 
8227 
1599 

•626 
4219 

i3i5 
4820 

8298 
1747 
5169 
8563 
1934 

4576 
8i36 
1667 
5i6g 
8644 
2091 
55io 
8903 
2270 

i347 
4934 
8490 
2018 
55j8 
8990 
2434 
585i 
9241 
26o5 

1707 
5291 
8845 
2370 
5866 
9335 

2777 
t»i9i 
9579 
2940 

2067 
5647 
9198 
2721- 
62i5 
9681 
3ii9 
653 1 

2426 
6004 
9552 
3071 
6562 
••26 
3462 
6871 
•253 
3609 

36o 
357 
355 
35i 

349 
346 
343 
340 
338 
335 

130 
131 
132 
133 
134 
135 
136 
137 
138 
139 

1 13943 
7271 

120D74 
3852 

TI05 

i3o334 
3539 
6721 

14301 5 

7603 
0903 
4178 
7429 
o655 
3858 
7037 
•194 
3327 

4611 

7934 

123l 

45o4 

7753 
0977 

7354 

•5o8 
3639 

4944 
8265 
i56o 
4830 
8076 
1298 
4496 
7671 
•822 
3951 

5278 
8595 
1888 
5i56 
8399 
1619 
4814 
7987 
1136 
4263 

56ii 

8926 
2216 
5481 
8722 

lilt 

83o3 
i45o 
4574 

5943 
9256 
2544 
58o6 
9045 
2260 
545i 
8618 
1763 
4885 

6276 
9586 
287I 
6i3i 
9368 
258o 
5769 
8934 
2076 
5196 

6608 

Zl 

6456 
9690 
2900 
6086 
9249 
2389 
5507 

6940 
•245 
3525 
6781 
••12 
3219 
64o3 
9564 
2702 
58i8 

333 
33o 
328 
325 
323 

321 

3i8 
3i5 
3i4 
3u 

140 
141 
142 
143 
144 
145 
146 
147 
148 
149 

146128 
9219 

152288 
5336 
83(2 

i6i3b8 
4353 
73,7 

170262 
3i86 

6438 
9527 
2594 
5640 
8664 
1667 
465o 
7613 
o555 
3478 

6748 
9835 
2900 
5943 
8965 
1967 

4947 
7908 
0848 
3769 

7o58 
•142 
32o5 
6246 
9266 
2266 
5244 
8203 
1141 
4060 

7367 

•449 
35io 
6549 
9567 
2564 
5541 
8497 
1434 
435i 

7676 
•756 
38i5 
6852 
9868 
2863 
5838 
8792 
1726 
4641 

7985 
io63 
4120 
7154 
•168 
3i6i 
6i34 
9086 
2019 
4932 

8:594 
1370 
4424 
7457 
•469 
3460 
643o 
9380 

23ll 
5222 

86o3 
1676 
4728 

7759 
•760 
3758 
6726 
9674 
2603 
55i2 

8911 
1982 
5o32 
8061 
1068 
4o55 

7022 

tl 

58o2 

309 
307 
3o5 
3o3 
3oi 
299 

293 
291 

150 
151 
152 
153 
154 
155 
156 
157 
158 
159 

176091 
8977 

181844 
4691 
7521 

190332 
3i25 
5900 
8657 

201397 

638i 
9264 
2129 

^vl 

0612 
3403 
6176 
8932 
1670 

6670 
9552 
24i5 
5259 
8084 

^] 

6453 
9206 
1943 

6959 
9b39 
2700 
5542 
8366 
1171 
3959 
6729 
9481 
2216 

7248 
•126 
2985 
5825 
8647 
i45i 
4237 
7oo5 
9755 
2488 

7536 
•41 3 
3270 
6108 
8928 
1730 
45i4 
7281 
..29 
2761 

7825 

6391 
9209 
2010 
4792 
,7556 
•3o3 
3o33 

8ii3 

•986 

ih 
6674 

9490 

228b 

5069 

7832 
•577 
31o5 

8401 
1272 
4123 
6956 

9771 
^^67 
5346 
8107 
•85o 
3577 

4407 
7239 
••5i 
2846 
5623 
8382 
1124 
3848 

289 
287 
285 

III 

-I 
It 

272 

N. 

0 

1   j  2 

3   14     5 

6  !  7 

8 

9 

D. 

LOGARITHMS   OF   NUMBERS. 


13 


N, 

0 

' 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

160 

204I20   4391 

4663 

4934 

5204 

5475 

5746 

6016 

6286 

6556 

271 

161 

6826   7096 

7365 

7634 

7904 

8173 

8441 

^V.9, 

8979 

9247  1  269 

162 

95i5  9783 

••5i 

•319 

•586 

•853 

1121 

i388 

1654 

1921  1  267 

163 

212188  2454 

2720 

2986 

3252 

35i8 

3783 

4049  i 

4814 

4379  266 

164 

4844  5i09  i  5373  i 

5638 

5902 

6166 

6430  j  6694 

6957 

9385 

7221   264 

165 

7484 

11^1 

8010 

8273 

8d36 

8798 

9060  1  9823 

9846  262 

166 

220108 

0370 

o63i 

0892 

ii53 

1414 

1675  ,  1986 

2196 

2456 

261 

167 

2716 

2976 

3236 

3496 

3755 

4oi5  4274  i  4533  1 

4792 

5o5i 

lU 

168 

5309 

5568 

5826 

6084  1  6342  ! 

6600 

6858  71 15 

7372 

7680 

169 

7887 

8144 

8400 

8657 

8913 

9170 

9426  9682 

9988 

•193 

256 

170 

230449 

0704 

0960 

12l5 

1470 

1724 

1979  2284 

2488 

2742 

254 

171 

2996 

325o 

35o4 

3757 

4011 

4264 

4317  4770 

5028 

5276 

233 

172 

5528 

5781 

6o33 

6285 

6537 

6789 

7041  7292 

7544 

7795 

252 

173 

^y  8046 

8297 

8548^ 

8799 

9049 
1D46 

9299 

9550  9800 

••30 

•800 

25o 

174 

240349 

3o38 

0799 

1(?48 

1297 

1795 

2044 

2298 

2541 

2790 

1$ 

175 

3286 

3534 

3782 

4o3o 

4277 

4525 

4772 

5019 

5266 

176 

55i3 

5759 

6006 

6252 

6499 

6745 

6991 

7287 

7482 

7728 

246 

177 

7973 

821? 

8464 

8709 

8954 
1395 

9198 

9443 

9687 

tts 

•176 

245 

173 

260420 

0664 

0908 
3338 

ii5i 

i638 

1881 

2125 

2610 

248 

179 

2853 

3096 

3580 

3822 

4064 

4806 

4548 

4790 

5o8i 

242 

180 

255273 

55i4 

5755 

5996 

6237 

6477 

6718 

6958 

7198 

7489 
9888 

241 

181 

7679 

7918 

8i58 

8398 

8637 

8877 

9116 

9855 

9594 

11^ 

182 

260071 

o3io 

0548 

0787 

1025 

1263 

i5oi 

1789 

1976 
4346 

2214 

183 

245i 

2688 

2925 

3i62 

3399 

3636 

8878 

4109 

4582 

287 

184 

4818 

5o54 

5290 

5523 

5761 

5996 

6282 

6467 

6702 

6987 

235 

185- 

7172 

7406 

7641 

7875 

8110 

8344 

8378 

8812 

9046 

9279 

284 

136 

95i3 

9746 

9980 

•2l3 

•446 

•679 

•912 

1 144 

1377 

1609 

283 

187 

271842 

2074 

23o6 

2538 

2770 

3ooi 

3233 

8464 

3696 

3927 

282 

188 

4i58 

4389 

4620 

485o 

5o8i 

53ti 

5542 

5772 

6002 

6232 

23o 

189 

6462 

6692 

6921 

7i5i 

7380 

7609 

7888 

8067 

8296 

8323 

229 

ll'O 

278754 

8982 

9211 

9439 

9667 

9895 

•123 

•85i 

•578 

•806 

228 

191 

281033 

1 261 

1488 

1715 

1942 

2169 

2896 

2622 

2849 

3075 

227 

192 

33oi 

3527 

3753 

3979 

42o5 

4431 

4656 

4882 

6107 

5882 

226 

193 

5557 

5782 

6007 

6232 

6456 

6681 

6905 

7180 

7354 

7578 

225 

194 

7802 

8026 

8249 

8473 

8696 

8920 

9143 

9866 

?^?? 

9812 

228 

195 

290035 

0257 

0480 

0702 

0925 

1147 
3363 

1869 

1591 

2084 

222 

196 

2256 

2478 

2699 

2920 

3i4i 

3584 

38o4 

4025 

4246 

221 

197 

4466 

4687 

4907 

5i27 

5347 

5367 

5787 

6007 

6226 

6446 

220 

198 

6665 

6884 

7104 

7323 

7542 

7761 

7979 

8.98 

8416 

8635 

^1^ 

199 

8853 

9071 

9289 

9507 

9725 

9943 

•161 

•878 

•595 

•8i3 

200 

3oio3o 

1247 

1464 

1681 

1898 

2114 

2881 

2547 

2764 

2980 

217 

201 

3196 

3412 

3628 

3844 

4039 

4275 

4491 

4706 

4921 

5i86 

216 

202 

53DI 

5566 

5781 

5996 

6211 

6425 

6639 

8778 

6854 

7068 

72S2 

2l5 

203 

7496 

7710 

7924 

8i37 

835i 

8564 

8991 

9204 

9417 

218 

204 

9630 

9843 

••56 

•268 

•481 

•693 

•906 

1118 

i33o 

1542 

212 

205 

311754 

1966 

2177 

2389 

2600 

2812 

3028 

3234 

3443 

3656 

211 

206 

3867 

4078 

4289 

4499 

4710 

4920 

5i3o 

5840 

5531 

5760 

210 

207 

5970 

6180 

6390 
8481 

6599 

6809 

7018 

7227 

7486 

7646 

7834 

209 

208 

8o63 

8272 

8689 

8898 

9106 

9314 

9522 

9780 

9938 

208 

269 

820146 

o354 

o562 

0769 

0977 

1184 

1891 

1598 

i8o5 

2012 

207 

210 

322219 

2426 

2633 

2839 

3046 

3252 

3458 

3665 

3871 

4077 

206 

211 

4282 

4488 

4694 

4899 
6950 

5io5 

53io 

55i6 

5721 

5926 

6181 

203 

£12 

6336 

6541 

6745 

7i55 

53^^ 

7563 

7767 

7972 

8176 

204 

213 

8380 

8583 

8787 

8991 

9194 

9601 

9805 

•••8 

•211 

203 

214 

33o4i4 

0617 

0819 

1022 

1225 

1427 

i63o 

1882 

2084 

2236 

202 

215 

2438 

2640 

2842 

3o44 

3246 

3447 

3649 

3850 

4031 

4253 

202 

216 

4454 

4655 

4856 

5o57 

5257 

5458 

5658 

5859 
7838 

6069 
8038 

6260 

201 

217 

6460 

6660 

6860 

7060 

7260 

7459 

7659 

8257 

200 

218 

8456 

8656 

8855 

9054 

9253 

9431 

965o 

9849 

••47 

•246 

198 

219 

340444 

0642 

0841 

1089 

1237 

I4e5 

I682 

1880 

2028 

2225 

N. 

1    " 

1 

2 

3 

4 

5 

i  ^ 

7 

8 

9 

D. 

K 

F 

14 


LOGARITHMS    OF    NU>rBERS. 


N. 

«    i  1 

2 

3 

.1   . 

6 

7 

8  1  , 

D. 

220 
221 
222 
228 
224 
225 
226 
227 
228 
229 

342423 

4392 
6353 
83o5 
350248 
2i83 
4108 
6026 
7935 
9835 

2620 

4589 
6549 
85oo 
0442 
2375 
43oi 
6217 
8125 

••25 

2817 
4783 
6744 
8694 
o636 
2568 
4493 
6408 
83i6 

•2l5 

3oi4 

4981 

til 

0829 
2761 

4685 
6599 
85o6 
•404 

3212 

5178 
7135 
9083 
1023 

2934 
4876 
6790 
8696 
•593 

3409 

5374 
7330 
9278 
1216 
3 147 
5o68 
6981 
8886 
•783 

36o6 
5570 

7523 

9472 
1410 
3339 
5260 

7172 
9076 
•972 

38oj 
5766 
7720 
9666 
i6o3 
3532 
5452 
7363 
9266 
1161 

3999 
5962 
7913 
9860 
1796 
3724 

5643 
7554 
9436 
i35o 

8110 
••54 

5834 
7744 
9646 
1539 

\^ 

195   1 

194 

193 

193 

192 

191 

\^ 

230 
231 
232 
233 
284 
235 
236 
237 
238 
239 

361728 
36i2 
5488 
7356 
9216 

371068 
2912 
4743 

IVol 
5675 
7342 
9401 
1253 
3096 
4932 
6759 
858o 

2io5 

7729 

9587 

1437 
3280 

5ii5 

6942 
8761 

2294 
4176 
6049 
7915 
9772 
1622 
3464 
5298 
7124 
8943 

2482 

4363 
6236 
8101 

??o6 
3647 
5481 
7306 
9124 

2671 
455i 
6423 
8287 
•143 

IK 

5664 
7488 
9306 

2859 
4739 
6610 
8473 
•328 
2173 

4013 

5846 
7670 
9487 

3048 
4926 
6796 
8639 
•5i3 
236o 
4198 
6029 
7852 
9668 

3236 
5ii3 
6983 
8845 
•698 

2344 

4382 
6212 
8o34 
9849 

3424 
53oi 
7.69 
9o3o 
•883 
2728 
4565 
6394 
8216 
••3o 

188 
188 
187 
186 

i85 
184 
184 
1 83 
182 
181 

240 
241 
242 
243  1 

244 ; 

245  ' 

246  , 
247 

24S  ; 

249  : 

380211 
2017 
38i5 
56o6 

I'Z 

390935 
2697 

4432 

6199 

0392 
2197 
3995 
5785 
7568 
9343 
1112 
2873 
4627 
6374 

0573 
2377 
4174 
5964 
7746 
9520 
1288 
3o48 
4802 
6548 

0754 
2557 
4353 
6142 

%i 

1464 

3224 

4977 

6722 

0934 
2737 
4533 
6321 
8101 
9875 
1641 
3400 
5i52 
6896 

III  5 

2917 
4712 

6499 
8279 
••5 1 

3373 
5326 

7071 

1296 
3097 
4891 
6677 
8456 
•228 

\% 
55oi 
7245 

1476 
3277 
3070 
6856 
8634 
•4o5 
2169 
3926 
5676 
7419 

1 656 
3456 
5249 
7034 
881 1 
•582 
2345 
4101 
585o 
7592 

1837 
3636 
5428 

vi 

2521 

4277 
6025 
7766 

181 
180 

-I 

178 

177 
176 
176 
175 
174 

250.' 
251 
252 
253 
254  '[ 

255 ; 

256 
257  , 
258 
259 

•397940 
9674 

40I40I 

3l2I 

4834 
6540 
8240 
9933 
41 1620 
3300 

8114 
9847 
1573 
3292 
5oo5 
6710 
8410 
•102 
1788 
3467 

8287 
••20 
1745 
3464 
5176 
688 1 
8579 
•271 

8461 

•192 
1917 

3633 
5346 
7o5i 

8749 
•440 
2124 
38o3 

8634 
•365 
2089 
3807 
55,7 
7221 
8918 
•609 
2293 
3970 

8808 
•538 
2261 
3978 
5683 
7391 
9087 

•777 
2461 
4i37 

8981 

•711 
2433 
4149 
5858 
756i 
9257 
•946 
2629 
4303 

9154 
•883 
26o5 
4320 
6029 
7731 
9426 
1114 
2796 
4472 

9328 
io56 

2777 
4492 
6199 

in 

1283 
2964 
4639 

9501 

1223 

294q' 

4663 

6370 

8070 

9764 

I45i 

3i32 

4806 

173 
173 
172 

171 
171 
170 
169 

167 

260 
261 
262  ' 
268  , 
264  ■. 

265 : 

265 
267 
268 
269 

414973 
6641 
83oi 
9956 

421604 
3246 
4882 
65ii 
8i35 
9752 

5i4o 
6807 
8467 
•121 
1768 
3410 
5045 
6674 
8297 
9914 

5307 
6973 
8633 
•286 
1933 
3574 
5208 
6836 
8459 
••75 

5474 

a 

•236 

5641 
73o6 
8964 
•616 
2261 
3901 
5534 
7161 
8783 
•398 

58o8 
7472 
9129 
•781 
2426 
4o65 
5697 
7324 
8944 
•559 

5974 
7638 
9295 

:^ 

4228 

5860 
7486 
9106 
•750 

6141 

7804 
9460 
mo 
2754 
4392 
6023 
7648 
9268 
•881 

63o8 
7970 
9625 
1275 
2918 
4555 
6186 
7811 
9^29 
1042 

6474 
8i35 

9791 
1439 
3082 
47>8 
6340 
7973 
9591 

I203 

167 
166 

1 65 
1 65 
164 
164 
1 63 
162 
162 
161 

270  1 

271 

£72 

278 

274 

275 

276 

277 

278 

279 

43i364 
2969 
4569 
6i63 
775i 
9333 

440909 
2480 
4045 
56o4 

i525 
3i3o 
4729 

6322 

7909 

'X 

2637 
4201 
5760 

1685 
3290 
4888 
6481 
8067 
9648 
1224 
2793 
4357 
5915 

1846 
345o 
5048 
6640 
8226 
9806 
i38i 
2950 
43i3 
6071 

2007 
36io 
5207 

3io6 
4669 
6226 

2167 
3770 
5367 
6957 
8542 
•122 
1695 
3263 
4825 
6382 

2328 
3930 

5D26 

7116 

8701 

•279 
1 852 
34T9 
49^1 
6537 

2488 
4090 
5685 
7275 
8839 
•437 
2009 
3576 
5i37 
6692 

2649 
4249 
5844 
7433 
9017 
•594 
2166 

3732 

3293 
6B48 

2S09 

4409 
6004 
7592 
9175 

•752 
2323 

3889 
5449 

7oo3 

161 
160 

ii 

i58 

\l^ 

1 56 
i55 

N. 

0 

1 

2 

3 

4    5 

6 

7 

8 

9 

D. 

LOGARITHMS    OF   NU:MBEE3. 


15 


K 

1   0  . 
1 

1 

2 

S 

4 

5 

G 

7 

8 

9 

D. 

280 
281 
282 
283 
284 
285 
286 
287 
288 
289 

447158 
8706 

450249 
1786 
33i8 
4845 
6366 
7882 
9392 

460898 

73i3 

8861 
o4o3 
1940 
3471 
4997 
65i8 
8o33 
9543 
1048 

7468 
9015 
0557 
2093 
3624 
5i5o 
6670 
8184 
9694 
1198 

7623 
9170 
07H 
2247 
3777 
53o2 
6821 
8336 
9B45 
1348 

7778 
9324 
o865 
2400 
3930 
5454 
6973 
8487 
9990 
1499 

7933 

9478 
101 3 
2553 
4082 
56o6 
7125 
8638 
•146 
1649 

8088 
9633 
1172 
2706 
4235 
5753 
7276 
8789 
•296 
1799 

8242 

9787 
i326 
2859 
4387 
5910 
7428 
6940 
•447 
1948 

8397 
9941 
1479 
0012 
4540 
6062 

7579 
9091 

2098 

8552 
••95 
i633 
3i65 
4692 
6214 
7731 
9242 
.743 
2243 

i55 

1 54 
1 54 
1 53 
1 53 

l52 
l52 

i5i 
i5i 
i5o 

290 
291 
292 
293 
294 
295 
296 
297 
298 
299 

462398 
3893 
5383 
6868 
8347 
9822 

471292 
2706 
4216 
5671 

2548 
4042 
5532 
7016 

8495 
9969 
1438 
2903 
4362 
58i6 

2697 

4191 
568o 
7164 
8643 
•116 
1 585 
3049 
45o3 
5962 

2847 
4340 
5829 
73.2 
8790 
•263 
1732 
3195 
46^3 
6107 

2997 
4490 
5977 
7460 
8933 
•410 
1873 
3341 
4799 

6232 

3i46 

4639 
6126 
7608 
Qo85 
•557 

2025 

3487 
4944 

6397 

3296 

4788 
6274 
7756 
9233 
•704 
2171 
3633 
5090 
6542 

3445 
4936 
6423 
7904 
9380 
•85i 
23i8 

i^v- 

5235 
6687 

6571 
8o52 
9527 
•998 
2464 
3925 
538i 
6832 

3744 
5234 
6719 
8200 
9675 
1145 
2610 
4071 
5526 
6976 

i5o 

149 
149 
143 
143 
147 
146 
146 
146 
145 

800 
801 
302 
803 
804 
805 

806  1 

807  ' 

808  1 
309 

477121 
8566 

480007 
1443 
2874 
43oo 
5721 
7i38 
855i 
9958 

7266 
8711 
oi5i 
1 586 
3oi6 
4442 
5863 
7280 
8692 
••99 

8855 
0294 
1729 
3i59 
4585 
6oo5 
7421 
8833 
•239 

7555 
&999 
0438 
1872 
33o2 
4727 
6147 
7563 
8974 
•38o 

7700 
9143 

o582 
2016 
3445 
4869 
6289 
7704 
9114 

•520 

7844 
9287 
0725 
2159 

3587 
Sou 
6430 

7845 

92DD 
•661 

7989 
943 1 
0869 

2302 
3730 

5id3 
6572 

7986' 

9396 

•801 

8i33 
9575 
1012 
2445 

3872 
5295 
6714 
8.27 
9537 
•941 

8278 

T.ll 

2588 
4oi5 
5437 
6855 
8269 

9677 
io8i 

8422 
9863 
1299 
2731 
4.57 
5579 
6997 
8410 
9818 
1222 

145 
144 
144 
143 
143 
142 
142 
141 
141 

■40 

810 
811 
812- 
813 
814 
815 
816 
817 
818 
819 

491362 
2760 
4i55 
5544 

9687 

D0I0D9 
2427 
3791 

l502 

2900 

4294 

5683 
7068 
8448 
9«24 
1196 
2564 
3927' 

1642 
3o4o 
4433 

5822 

7206 
8586 

??M 

2700 
4o63 

1782 
3i79 
4572 
5q6o 
7344 
8724 
••99 
14-^ 
2837 

4199 

1922 
3319 
47II 
6099 

7483 
8862 
•236 
1607 
2973 
4335 

2062 

3458 
485o 
6238 
7621 

1744 
3109 
4471 

2201 

4989 
6376 

7759 
9137 
•5u 
1880 
3246 
46©7 

2341 

3737 
5i23 
65i5 

7897 
9273 
•643 
2017 
3382 
4743 

2481 

3876 
5267 
6653 
8o35 
9412 
•785 

2ID4 

35i8 

4878 

2621' 
4oi5 
5406 
6791 
8173 
9550 
•922 
2291 
3653 
5oi4 

140 
139 
139 
139 
i33 
1 33 
137 
137 
i36 
1 36 

820 
821 
822 
823 
824 
825 
826 
327 
823 
^  829 

5o5i5o 
65o5 
7856 
9203 

5io545 
1 883 
3218 
4548 
5874 
7196 

5286 
6640 

7991 
9337 
0679 
2017 
335i 
4681 
6006 
7328 

5421 

6776 
8126 

9471 
o8i3 

2l5l 

3484 
48i3 
6139 
7460 

5557 
691 1 
8260 
9606 
0947 
2284 
3617 
4946 
6271 
7592 

5693 
7046 
8395 
9740 
1081 
24.8 
3750 
5o79 
64o3 
7724 

5828 
7181 
853o 
9874 

12l5 

255i 
3883 

52II 

6535 
7855 

5964 
7316 
8664 
•••9 
1 349 
2684 
4016 
5344 
6668 
7987 

6099 
7431 

1482 
2818 
4149 
5476 
6800 
8119 

6234 
7586 
8934 
•277 
i6i6 
2951 
42S2 
5609 
6932 
825i 

6370 
7721 
9063 
•411 
1750 
3oS4 
44i5 
5741 
7064 
8382 

136 
i35 
i33 
1 34 
i34 
i33 
i33 
i33 

l32 
l32 

830 
331 
332 
333 
834 
835 
336 
337 
333 
839 

N. 

5i85i4 
9823 

52II38 
2444 
3746 
5o45 
6339 
763o 

530200 

8646 
9959 
1269 
2575 
3876 
5i74 
6469 
7759 
9043 
0328 

8777 
••90 
1400 
2705 
4006 
53o4 
6598 
7888 
9174 
0456 

8909 
•221 
i53o 
2835 
4i36 
5434 
6727 
8016 
9302 
o584 

9040 
•353 
1661 
2966 
4266 
5563 
6856 
8145 
94  3  0 
0712 

9171 
•484 
1792 
3096 
4396 

6985 
8274 
9559 
0840 

93o3 
•6i5 
1922 
3226 
4526 

5822 

8402 

9434 
•745 
2053 
3356 
4656 
5951 
7243 
853 1 
9815 
1096 

9566 
•376 
2i83 
3486 
4785 
6081 
7372 
8660 

9697 
1007 
23 14 
36i6 
49'5 
6210 
75oi 
8788 
••72 
i35i 

i3i 
i3i 
i3i 
i3o 
i3o 
129 
129 

III 

128 

0 

1 

2 

3 

4 

5 

6 

^ 

' 

9 

D. 

16 


LOGARITHMS    OF    NUMBERS. 


N. 

1  0 

1 

2 

8 

4 

5 

6 

7  1  8-    9  1  D.  1 

S40 

531479  !  1607  i  1734 

1862 

1990  j  2117 

2245  1  2372  j  25oo 

2627 

128 

841 

2754  2882  i  3009 

3i36 

3264  1  3391 

35i8  I  3645  !  3772 

3899 

127 

842 

4026  41 53  '  4280 

4407 

4534  4661 

4787  U914  !  5o4i 

5167 

III 

848 

5294  5421  :  5547 

5674 

58oo  5927 

6o53  1  6180  1  63o6 

6432 

844 

65o8  ;  6685  68  u 

6937 

7063 

7189 

8574  '  8699  ^823 

^] 

126 

845 

7819  7945  8071 

8197 

8322 

8448 

126 

846 

9076  ,  9202  9327 

9432 

9578 

9703 

9829  9934  ••79 

•204 

125 

847 

540329  1  0455  od8o 

0705 

o83o 

0955 

io3o  ;  i2o5  i33o 

1454 

125 

848 

1679  1704  1  1829 

1953 

2078 

2203 

2327  2452  2576 

2701 

123 

349 

2823  2930  1  3074 

3199 

3323 

3447 

3571  !  3696  3820 

3944 

124 

850 

•  544068  1  4192 

43 16 

4440 

4564 

46S8 

4812  4936  1  5o6o 

5i83 

124 

851 

!   5307  !  543 1 

5555 

5078 

58o2 

5925 

6049  6172  1  6296 

6419 

124 

852 

6543 

6666 

6789 

6913 

7o36 

7.59 

72S2  {  7403 

3,1? 

7652 

123 

858 

:  7773 

7S93 

802 1 

8144 

8267 

8389 

85i2  1  8635 

8881 

123 

854 

9003 

9126 

9249 

9371 

9494 

9616 

9739  !  9861 

9984 

•106 

123 

855 

5502 28 

o35i 

047-i 

0595 

0717 

0840 

0962  1  10S4 

1206 

i328 

122 

856 

1430 

1572 

1694 

1816 

1933 

2060 

2181  j  23o3 

2423 

2547 

122 

857 

2663 

2790 

2911 

3o33 

3i55 

3276 

3398  1  3519 

3640 

3762 

121 

858 

'   3583 

4004 

4126 

4247 

4368 

44&9 

4610  ;  4731 

4852 

4973 

121 

S59 
860 

:  5094 

52i5 

5336 

5457 

5573 

5699 

5820  5940 

6061 

6I82 

121 

5563o3 

6423 

6544 

6664 

6785 

6905 

7026  i  7146  i  7267 

7387 

120 

861 

7507 

7627 

7748 
8948 

7868 

7988 

8108 

8228  1  8349  i  8469 

858^ 

120 

862 

8709 

8829 

9068 

9188 

9308 

9428  '  9548  j  9667 

9787 

120 

863 

9907 

••26 

•146 

•265 

•385 

•5o4 

•624  !  ^743  i  ^863 

•982 

119 

864 

56IIOI 

1221 

1 340 

1459 

1578 

1698 

1817  ;  1936 

2055 

2174 

119 

365 

2293 

2412 

253 1 

265o 

2769 

2887 

3oo6  i  3i25 

3244 

3362 

119 

SC6 

3481 

3600  1  3718 

3837 

3955 

4074 

4192  i  43ii 

4429 

4548 

\;i 

367 

4666 

4784 

4903 

502I 

5i39 

5257 

5376 

5494 

5612 

5730 

868 

5848 

5966 

6084 

6202 

6320 

6437 

6555 

6673 

6791 

6909 

118 

869 

7026 

7144 

7262 

7379 

7497 

7614 

7732 

7849 

7967 

8084 

118 

870 

568202 

83i9 

8436 

8554 

8671 

8788 

8905 

9023 

9140 

9257 

117. 

871 

.9374 

949 » 

9608 

.9725 

9842 

9939 

••76 

•193 

•309 

•426 

117 

872 

570043 

0660 

0776 

0893 

lOIO 

1126 

1243 

1359 

1476 

1392 

"I 
116 

873 

1709 

1S25 

1942 

2058 

2174 

2291 

2407 

2323 

2639 

2755 

374 

2872 

2988 

3104 

3220 

3336 

3432 

3568 

3684 

3800 

39.3 

116 

S75 

4o3i 

4147 

4263 

4379 

5534 

4494 

4610 

4726 

4841 

4957 

5672 

116 

376 

5.88 

53o3 

5419 

5630 

5765 

588o 

5996 

6111 

6226 

ii5 

877 

6341 

6457 

6572 

6687 

6802 

6917 

7o32 

^47 
8293 

7262 

7377 

ii5 

378 

7492 

7607 
8754 

mi 

7836 

7951 

8066 

8181 

8410 

8525 

ii5 

879 

8639 

8983 

9097 

9212 

9326 

9441 

9555 

9669 

114 

880 

579784 

9898 

••12 

•126 

•241 

•355 

•469 

•583 

•697 

•811 

114 

881* 

580925 

1039 

ii53 

1267 

i38i 

1495 
263 1 

1608 

1722 

1836 

1950 

114 

882 

2063 

2177 

2291 

2404 

25i8 

2745 

2858 

2972 

3o85 

114 

883 

3199 

33i2 

3426 

3539 

3652 

3765 

3879 

3992 

4103 

4218 

ii3 

884 

433 1 

4444 

4557 

4670 

4:83 

4896 

3009   5l22 

5235 

5348 

ii3 

885 

5461 

5574 

5686 

5799 

5912 

6024 

6137 

62  5o 

6362 

6475 

ii3 

886 
887 

6587 

nil 

6700 
7823 

6812 
7935 

6925 
8047 

7037 
8160 

7149 
8272 

7262 

8384 

c 

7486 
8608 

7399 
8720 

112 
112 

888 

8944 

9o56 

9167 

9279 

9391 

95o3 

90.5 

9726 

9838 

112 

889 

9950 

••61 

•173 

•284 

•396 

•307 

•619 

.-,30 

•842 

•953 

112 

890 

591065 

1176 

2288 

1287 

1399 

i5io 

1621 

1732   1843 

1935  2066 

III 

891 

2177 

2399 

25lO 

2621 

2732 

2843  ■   2954 

3o64 

3175 
4282 

III 

892  1 

3286 

3397 

35o8 

36i8 

3729 

3840 

3950  ]  4061 

4171 

III 

393  1 

4393 

45o3 

4614 

5827 

4834 

4945 

5o55  I  5i65 

5276 

5386 

no 

894  ! 

5496 

56o6 

5717 

5937 

6047 

61 57  i  6267  6377 

6487 

no 

395  , 

6597 

769D 

6707 

6817 

6927 

i:li 

7146 

7256  i  7366  i  7476 

8681 

no 

306  t 

7805 
8900 

7914 

8024 

8243 

8353  1  8462  8572 

no 

S97  ' 

9009 

9119 

9228 

9337 

9446  1  9556  !  966J 

9774 

109 

393 

9883  9992 
600973  I  1082 

•lOI 

•210  1  •319  •428  •537  '646  1  •755 

•864 

109 

899  ■ 

1191 

1299  1  1408  i5i7  1625  1734  1  1843 

,951 

109 

N.  1 

0     1 

2 

3   1  4  1  5  1   6   1  7     8 

9 

D. 

LOGARITHMS    OF   NUMBERS. 


17 


N.  1   0 

I 

2 

3 

4 

5 

6 

7 

8 

9    D. 

400  i 

401  i 

402  , 
403 
404 
405 
406 
407 
403 
409 

602060 
3i44 
4226 
53o5 
6331 
7455 
8526 
^  9594 
610660 
1723 

2169 
325J 
4334 
54i3 
6489 
7562 
8633 
9701 
0767 
1829 

2277 
336i 
4442 
5521 
6596 
7669 
8740 
9808 
0873 
1936 

2386 
3469 
455o 
5628 
6704 

^847 
9914 
0979 
2042 

2494 
3577 
4658 
5736 
6811 
7884 
8954 
••21 
1086 
2148 

2603 
3686 
4766 
5844 
6919 
7991 
9061 
•128 
1192 
2234 

2711 

3794 
4874 
5951 
7026 
8098 
9167 
•234 
1298 
2360 

2819 
3902 
4982 
6059 
7.33 

8203 

2S 

i4o5 
2466 

2928 
4010 
5089 
6166 
7241 
8312 
9381 
•447 
i5ii 
2572 

3o36 
4118 
5197 
6274 
7348 
8419 
94b8 
•554 
1617 
2678 

108 

lOb 

108 
108 
107 
107 
107 
107 
106 
106 

410 
411 
412 
413 
414 
415 
416 
417 
41S 
419 

612784 
3842 

tVo 
7000 
8048 
9093 
620136 
1176 
2214 

2890 

llU 

6o55 
7105 
8i53 
9198 
0240 
1280 
23i8 

lit, 

5io8 
6160 
7210 
8257 
9302 
o344 
1 384 

2421 

3l02 

4139 
52i3 
6265 
73i5 
8362 
9406 
0448 
1488 

2525 

32o7 
4264 
5319 
6370 
7420 
8466 
9311 
o552 
1592 
2628 

33 13 
4370 
5424 
6476 
7525 
8571 
9615 
0656 
1695 
2732 

3419 
4473 
5529 
658i 

9719 
0760 
1799 
2835 

3525 

458 1 
5634 
6686 
7734 
8780 
9824 
0864 
1903 
2939 

363o 

4686 
5740 
6790 

8884 
9928 
0968 
2007 
3o42 

3736 
4792 
5»45 
6895 
7943 
89«9 

••32 

1072 
2110 
3146 

106 
106 
I05 
103 

io5 
io5 
104 
104 
104 
104 

420 
421 
422 
423 
424 
425 
426 
427 
4£8 
429 

623249 
4282 
53i2 
6340 
7366 
8389 
9410 

630428 
1444 
2457 

3353 
4385 
541 5 
6443 
7468 
8491 
9512 
o53o 
i545 
2559 

3456 

4488 
55i8 
6546 

llll 

9613 
o63i 
1647 
2660 

3559 
4591 

5621 
6648 
7673 
8695 
9715 
0733 
1748 
2761 

3663 
4695 
5724 
6731 
7773 
8797 

^^ 
1849 
2802 

3766 

4798 
5827 
6853 
7878 
8900 
9919 
0936 
1931 
2963 

3869 
4901 

l$t 

7980 
9002 
••21 
io38 

2052 

3o64 

3973 
5oo4 
6o32 
7o58 
8082 
9104 

•123 

1139 
21 53 
3i65 

4076 
5107 
6i35 
7161 
8i85 
9206 
•224 
1241 

2255 

3266 

4179 

52IO 

6238 
7263 
8287 
9308 
•326 
i342 
2356 
3367 

io3 
io3 

io3 
io3 
102 
102 
102 
102 

lOI 

101 

430 
431 
432 
433 
4^4 
435 
436 
4G7 
438 
439 

633468 
6488 

IX 

9486 

640481 

1474 

2465 

3569 

437^ 
5584 
6588 

9586 
o58i 

2563 

3670 
4679 
5685 
6688 
7690 
8689 
9686 
0680 
1672 
2662 

3771 
4779 
3785 
6789 

7790 
^789 
9783 
0779 
1771 
2761 

3872 
4880 
5886 
6889 
7890 
8888 
9885 
0879 
1871 
2860 

3973 
4981 
5986 
6989 
7990 
8968 
9984 
0978 
1970 
2939 

4074 
5o8i 
6087 
7089 
8090 
9088 
••84 
1077 
2069 
3o58 

4175 
5182 
6187 
7.89 
8190 
9188 
•i83 
1177 
2168 
3i56 

4276 

5283 
6287 
7290 
8290 
9287 
•283 
1276 
2267 
3255 

4376 
5J8J 
6388 
7390 
8389 

93^7 
•382 
1375 
2366 
3354 

100 
100 
100 
100 
99 
99 
99 
99 
99 
99 

440 
441 
442 
443 
444 
445 
446 
447 
448 
*449 

643453 
4439 
5422 
6404 
7383 
836o 
9335 

65o3o8 
1278 
2246 

355i 
4537 
552i 
65o2 
7481 
8458 
9432 
o4o5 
1375 
2343 

365o 
4636 
56i9 
6600 

7579 

85DD 

9530 

o5o2 

1472 
2440 

3749 
4734 
5717 
6698 
7676 
8653 
9627 
0599 
1369 
2536 

3847 

4832 

58i5 

6796, 

7774 

8750 

9724 

0696 

1666 

2633 

3946 
4931 
59.3 
6894 

& 

9821 
0793 
1762 
2730 

4044 
5029 
6011 
6992 
7969 
8943 
9919 
0890 
1839 
2826 

4143 
5127 
6110 
7089 
8067 
9043 
••16 

tl 

2923 

4242 
5226 
6208 

v:ii 

9140 

•ii3 

1084 
2o53 
30.9 

4340 
5324 
63o6 
7285 
8262 
9237 
•210 
11^1 

2l50 

3ii6 

98 

9.^ 

9^ 
98 
98 

97 
97 
97 
97 
97 

450 
451 
452 
453 
454 
455 
456 
467 
458 
459 

653213 

mi 

6098 
70D6 
8011 
8965 
9916 
660865 
i8i3 

3309 
4273 
5235 
6194 

7102 
8107 
9060 
**II 
0960 
1907 

34o5 
4369 
5331 
6290 
7247 
8202 
9155 
•106 
io55 
2002 

35o2 
4465 
5427 
6386 
7343 
8298 
9230 
•201 
ii5o 
2096 

3598 
4562 
5523 
6482 
7438 
8393 
9346 
•296 
1245 
2191 

3693 
4658 
5619 
6577 
7534 
8488 
9441 
•391 
1339 
2286 

3791 
4734 
5715 
6673 

til 

9536 
•486 
1434 
238o 

3888 
485o 
58io 
6769 

7723 

8679 
9631 

•58 1 
i529 
2473 

3984 
4946 
5906 
6864 
7820 
8774 

1623 

2569 

40S0 
5o42 
6002 
6960 
7916 
8870 
9821 

1718 
2663 

96 
96 
9b 
96 

96 

9D 

9? 
9? 
93 

N. 

1   ^ 

1 

2 

3 

* 

5 

« 

' 

8 

9   1  D. 

18 


LOGARITHMS    OF   l^UMBERS. 


N.  1;   0 

1 

2 

a 

4 

0 

6 

7 

8 

35i2 
4454 

lit 

7266 

8199 
9i3i 
..60 
0988 
1913 

9   i  D.  1 

4  SO 
461 
462 
463 
464 
465 
466 
467 
463 
469 

662753 

!  3701 
■i     4642 
1   55«i 
1   65i8 
1453 
8386 

1  670246 
1   1173 

2852 

3795 

4736 
5675 
6612 
7546 

8479 
9410 
0339 
1265 

I  2947 
3»89 
483o 

7640 
8J72 
95o3 
043 1 
i358 

3o4i 

3983 

4924 

5ttd2 

^fs 

8665 
9596 

0024 

I45i 

3i35 

4078 
5oi8 
5956 
6892 
7826 
8759 
9689 
0617 
i543 

323o 
4172 

5ll2 

6o5o 
6986 
7920 
8852 
9782 
0710 
1636 

3324 
4266 
5206 
6143 
7079 
8oi3 
8945 
9875 
0802 
1728 

3418 
436o 

7173 
bio6 
9o38 

tl 

1821 

4548 
5487 
6424 

8293 
9224 
.153 
1080 
2oo5 

94 
94 
94 

93 

470 

471 
472 
473 

474 
475 
476 

477 
47S 
479 

480 
451 
432 
453 
484 
455 
486 
487 
433 
4S9 

1  672098 

302I 

3942 
4861 

5778 
6694 

n 

1    9428 

:  68o336 

2190 
3ii3 
4o34 
4953 
5870 
6785 
7698 
8609 
9319 
0426 

2283 

32o5 
4126 
5o45 
5962 
6876 

7789 
8700 
9610 
o5i7 

2375 
3297 
4218 
5i37 
6o53 
6968 
7881 
8791 
9700 
0607 

2467 
3390 
43io 
5228 
6145 
7059 
7972 
8»82 

9791 
0698 

256o 
3482 
4402 
5320 
6236 
7i5i 
8o63 
8973 
9882 
0789 

2652 

3574 

4494 
5412 

6328 
7242 
8i54 
0064 
9973 
0879 

2744 
3666 
4586 
55o3 

6419 
7333 
8245 
9155 
..63 
0970 

2836 
3753 

65ii 
7424 
8336 
9246 
•154 
1060 

UP, 

6602 
7516 
8427 

ii5i 

92 
92 
92 
92 
92 
91 
91 
91 
91 
91 

681241 
2145 
1   ^°47 
3947 
4^45 
5742 
6636 
7529 
8420 
9309 

i332 

2235 

3i37 
4037 
4935 
583i 
6726 

r, 
9393 

1422 

232b 

3227 
4127 
5o25 
5921 
681 5 

mi 
9486 

i5i3 
2416 
3317 
4217 
5ii4 
6010 
6904 
7796 
8087 
9375 

i6o3 
25o6 
3407 
4307 
5204 
6100 
6994 
7886 
8776 
9664 

1693 
2596 

3497 
4396 
5294 
6189 
7083 
7975 
8865 
9753 

1784 
2686 
3587 
4486 
5383 
6279 
7172 
8064 
8953 
9»4i 

1874 
2777 
3677 
4376 
5473 
6368 
7261 
8i53 
9042 
9930 

1964 
2867 
3767 
4666 
5563 
6458 
735i 
8242 
9i3i 
..19 

2o55 
2957 
3857 
4756 
5652 
6547 
7440 
8331 
9220 
•107 

90 
90 
90 
90 

II 

89 

89 

490 
491 
4y2 
493 
494 
495 
496 
4y7 
493 
4y9 

690196 

1081 
1965 
2847 
3727 
46o5 
5482 
6356 
7229 
8101 

0285 
1 1 70 
2o53 
2935 
3Si5 
4693 
5569 
6444 
7317 
8i83 

0373 
1258 
21-42 
3o23 
3903 
4781 
5657 
653 1 
7404 
8275 

0462 
i347 

223o 

3iu 
3991 
4863 
5744 
6618 

7491 
8362 

o55o 
1435 
23i8 
3199 
4078 
4956 
5&32 
6706 
7578 
8449 

0639 
i524 
2406 

3287 
4166 
5o44 

7665 
8535 

0728 
1612 
2494 
3375 

4234 

5i3i 
6007 

6880 

7732 

8622 

0816 
1700 

2353 

3463 
4342 
5219 
6094 
6968 
7839 
8709 

0905 

1789 
2671 
355i 
443o 
5307 
6182 
7055 

^^ 

0993 

1877 

nil 

4517 
5394 
6269 

8014 

8883 

89 
88 
88 
83 
83 
88 
87 

?7 
87 
87 

500 
501 
502 
503 
504 
505 
506 
507 
503 
509 

700704 
1 568 
243 1 
3291 

5oo8 
5864 
6718 

9057 
9924 
0790 
i654 
25i7 

4236 
5094 
5949 
68o3 

9144 
••11 

0877 
1741 
26o3 
3463 

4322 

I'oll 

68tt8 

923i 
••98 
0963 
1827 
26S9 
3549 
4408 
5265 
6120 
6974 

9317 
•184 
io5o 
19.3 
2773 
3635 
4494 
5350 
6206 
7059 

9404 
•271 
ii36 
1999 
2861 
3721 

4579 
5436 
6291 
7»44 

9491 

•358 
1222 

2086 
2947 
3807 
4665 

5522 

6376 
7229 

9578 
.444 
i3o9 
2172 
3o33 

^??? 

5607 
6462 
73i5 

1395 

2258 

3119 

3979 

4837 
5693 
6547 
7400 

9751 
.617 
1482 
2344 
32o5 
4o65 
4922 
5778 
6632 
7485 

87 
87 
86 
86 
86 
86 
86 
86 
85 
85 

510  i!  707570 

511  i:  8421 

512  1^   9270 

512  1,  710117 

514  0963 

515  i'   1807 

516  !   265o 

517  ■       3491 

513  ■   433o 
519  :   5167 

7655 
85o6 
9355 
0202 
1043 

IX 

3575 
4414 

525i 

7740 
8591 
9440 
0287 

Il32 

1976 
2H18 
3659 

4497 
5335 

7826 
8676 
9324 
0371 
1217 
2060 
2902 
3742 
458 1 
5418 

9609 

0456 
i3oi 
2144 
2986 
3826 
4665 
55o2 

7996  8081 
8846  8931 

9694  ;  9779 

o54o  0620 
i385  1470 
2229  23i3 
3070  3 1 54 
3910  1  3994 
4749  i  4833 
5586  1  5669 

8166 
90i5 
9S63 
0710 
1 554 
2397 
3238 
4078 
4916 
5733 

825i 
9100 
9948 
0794 
1639 
2481 
3323 
4162 
5ooo 
5836 

8336 

??]? 

2566 
3407 
4246 
5o84 
5920 

85 

85 
85 
85 
84 
84 
84 
84 
84 
84 

-X. 

0 

1   i  2 

3     4 

5   1   6 

7 

8 

9 

D. 

LOGARITBLMS   OF   NUMBERS. 


19 


N. 

i   0     1 

2 

3 

4 

5 

6 

7 

8  i  9 

I). 

520 
521 
522 
523 
524 
525 
526 
527 
528 
529 

716003 
6838 
7671 
85o2 
9331 

;  720109 
1    0986 
i    181I 

j   2634 
i   3456 

6087 
6921 

lili 

9414 
0242 
1068 
1893 
2716 

3538 

6170 
7004 
7837 
8668 
9497 
o325 
ii5i 
1975 
2798 
3620 

6254 
70S8 
7920 
8751 
9580 
0407 
1233 
2o58 
2881 
3702 

6337 

8834 
9663 
0490 
i3i6 
2140 
2963 
3784 

6421 
7254 
8086 
8917 
9745 
0573 
1398 
2222 
3o45 
3866 

65o4 
7338 
8169 
9000 
9828 
0655 
1481 
23o5 
3.27 
3948 

6588 
7421 
8253 
9083 
9911 

:r^ 

2387 
3209 
4o3o 

6671 
7504 

8336 
9165 
9994 
0821 
1646 
2469 
3291 
4112 

6754 

7587 
8419 
9248 

••77 
0903 
1728 

2552 

3374 
4194 

83 
83 
83 
83 
83 
83 
82 
82 
82 
82 

530 
531 
532 
533 
534 
535 
536 
537 
538 
539 

i  724276 

;   5095 

591-2 

6727 

8354 
9165 
9974 

4358 
5176 
5993 
6809 
7623 
8435 
9246 
••55 
o863 
1669 

4440 
5258 
6075 
6890 
7704 
85i6 
9327 
•i36 
0944 
1750 

4522 

5340 

61 56 
6972 
7785 
8597 
9408 
•217 
1024^ 
i83o 

46o>f 
5422 
6238 
7053 
7866 
8678 

iio5 
191 1 

4685 
55o3 
6320 
7134 
7948 
8739 
9370 
•378 
1186 
1991 

4767 
5585 
6401 
7216 
8029 
8841 
965 1 
•459 
1266 
2072 

4849 
5667 
6483 

7297 
8uo 
8922 
9732 
•540 
1347 

2l52 

4931 
5748 
6564 

7379 
8191 
9003 
9813 
•621 
1428 

2233 

5oi3 
583o 
6646 
7460 
8273 
9084 
9893 

:s 

23i3 

82 
82 
82 
81 
81 
81 
81 
81 
81 
81 

540 
541 
542 
543 
.544 
545 
546 
547 
548 
549 

732394 

3999 
4800 

?;§^ 

8781 
9572 

2474 
3278 
4079 

4880 
5679 
6476 
7272 
8067 
8860 
965 1 

2555 
3358 
4160 
4960 
5759 
6556 
7352 
8146 
8939 
9731 

2635 
3438 
4240 
5o4o 
5838 
6635 
7431 
S225 
9018 
9S10 

2715 
35i8 
4320 

5l20 

5918 
6715 
75ii 
83o5 
9097 
9889 

2796 
3598 
4400 

5200 

5998 

6795 
7590 

8384 

2876 
3679 
4480 
5279 
6078 
6874 
7670 
8463 
9256 
••47 

2956 
3759 
456o 
5359 
6157 
6934 
7749 
8543 
9335 
•126 

3o37 
3839 
4640 
5439 
6237 
7034 
7829 
8622 
9414 

•205 

3.17 
3919 
4720 
5319 
63i7 
7113 
7908 
8701 
9493 
•284 

80 

80 
80 
80 
80 
80 
79 
79 
79 
79 

550 
551 
552 
553 
554 
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556 
557 
558 
559 

740363 

Il52 

Itt 

35io 
4293 
5o73 
5855 
6634 
7412 

0442 

I230 

2018 
2804 

3588 
4371 
5i53 
5933 
6712 
7489 

052I 

1 309 
2096 

2882 
3667 
4449 

523i 
601 1 
6790 
7567 

0600 

i388 

2961 
3745 
4528 
53o9 

6868 
7645 

0678 
1467 
2254 
3o39 
3823 
4606 
5387 
6167 
6945 
7722 

0757 
1 546 

2332 

3ii8 
3902 
4684 
5465 
6245 
7023 
7800 

o836 
1624 
241 1 
3196 
3980 
4762 
5543 
6323 
7101 
7878 

0915 
1703 
2489 
3275 
4o58 
4840 

5021 

6401 

0994 

111 

3353 
4i36 

4919 
5699 

6479 
7256 
8o33 

1073 
i860 
2647 
343 1 

4213 

4997 

5777 

6556 
7334 
8110 

79 

79 
79 

]t 

78 
7^ 
78 
78 
78 

560 
561 
562 
563 
564 
565 
566 
567 
568 
'   569 

748188 
8963 

75^508 
i?79 
2048 
2816 
3583 
•4348 

5lI2 

8266 
9040 
9814 
o586 
1 356 

2125 

2893 

3660 
4425 
5189 

8343 
9118 
9891 
o663 
1433 
2202 
2970 
3736 

4301 

5265 

8421 
9195 
9968 
0740 
i5io 
2279 
3o47 
38i3 
4578 
5341 

8498 
9272 
••45 
0817 
1 587 
2356 
3i23 
3889 
4654 
5417 

8576 
9350 

•I23 

0894 
1664 
2433 

3200 

3966 
4730 
5494 

8653 
9427 
•200 
0971 

2  509 

3277 
4042 
4807 
5570 

8731 
9504 
•277 
1048 
1818 
2586 
3353 
4119 
4883 
5646 

8808 
9582 
•354 

II25 

1895 

2663 
343o 
4195 
4960 
5722 

8885 
9659 
•43 1 
1202 
1972 
2740 
35o6 
4272 
5o36 
5799 

77 
77 
77 
77 
77 
77 
77 
77 

^t 
76 

570 
571 
572 
573 
574 
575 
576 
577 
578 
579 

755875 
6636 

lit. 

8912 
9668 
760422 
1 1 76 
1928 
2679 

595. 
6712 

7472 
823o 
8988 
9743 
0498 

I25l 
2003 

2754 

6027 

6788 
7548 
83o6 
9063 

??;? 

i326 
2078 
2829 

6io3 

6864 
7624 
8382 
9139 
9894 
0649 
1402 
2i53 
2904 

6180 
6940 

7700 
8458 
9214 
9970 
0724 
1477 
2228 
2978 

6256 

7016 
7775 
8533 
9290 
••45 
0799 
i552 
23o3 
3o53 

6332 
7092 
7831 
8609 
9366 
•121 
0875 
1627 
2378 
3i:.8 

6408 
7168 

7927 
8685 
9441 
•196 
0930 
1702 
2453 
32o3 

6484 
7244 
8oo3 
8761 
9517 
•272 

1023 

1778 
2329 
3278 

656o 
7320 

8079 
8836 
9592 
•347 

IIOI 

i853 
2604 
3353 

76 
76 
76 
76 
76 

?5 

75 

75 

75 

N. 

0 

1 

2 

3  J  . 

5 

6 

7 

8     9   1 

D. 

20 


LOGARITHMS    OF   KUMBEES. 


N.  \ 

0 

1 

2 

8 

4 

5 

6 

7 

8 

9 

D. 

580  ; 

763428 

35o3 

3578 

3653 

3727 

38o2 

3877 

3952 

4027 

4101 

i 

5S1 

4176 

425i 

4326 

4400 

4475 

455o  4624  1 

4699 

4774 

4848 

7? 

582  i 

4923 

4998 

5072 

5i47 

5221 

5296 

5370 

5445 

5520 

5594 

75 

583  1 

If^ 

5743 

58i8 

5892 

5966 

6041 

6ii5 

6190 

6264 

6338 

74 

584  1 

6487 

6562 

6636 

6710 

6785 

6859 

6933 

7007 

7°^^ 

74 

585 

7i56 

723o 

i3o4 
8046 

7379 

J453 

7527 

7601 

7675 
8416 

7749 

7823 
8564 

74 

586 

lt& 

7972 

8120 

8268 

8342 

8490 

74 

637 

8712 

87S6 

8860 

8934 

9008 

9082 

91 56 

9230 

93o3 

74 

5S8 

9^77 

9451 

9525 

9599 

9673 

9746 

9820 

9894 

9968 

••42 

74 

589 

770u5 

0189 

0263 

o336 

0410 

0484 

0557 

o63i 

0705 

0778 

74 

590 

770802 

0926 

0999 

1073 

1 1 46* 

1220 

1293 

1 367 

1440 

i5i4 

74 

591 

i587 

1661 

1734 

1808 

1881 

1955 

2028 

2102 

2175 

2248 

73 

592 

2322 

2395 

2468 

2542 

26i5 

2688 

2762 

2835 

2908 

2981 

73 

593 

3o55 

3128 

3201 

3274 

3348 

3421 

3494 

3567 
4298 

3640 

3713 

73 

594 

3-86 

3860 

3933 

4006 

4079 

4132 

4225 

4371 

4444 

73 

595 

4317 

4590 

4663 

4736 

4809 
5538 

4882 

4955 

5028 

5ioo 

5173 

73 

596 

5246 

5319 

5392 

5465 

56io 

5683 

5756 

5829 

5902 

73 

597 

5974 

6047 

6120 

6193 

6265 

6338 

641 1 

6483 

6556 

6629 

*?? 

598 

6701 

6774 

6846 

6919 

6992 

7064 

7'j7 

7209 

7282 
8006 

7334 
8079 

73 

599 

7427 

7499 

7372 

7644 

7717 

7789 

7862 

7934 

72 

600 

778131 

8224 

8296 

8368 

8441 

85i3 

8585 

8658 

8730 

8802 

72 

601 

'8874 

8947 

9019 

9091 

9'^^ 

9236 

9308 

9380 

9452 

9524 

72 

602 

9596 

9669 

9741 

98.3 

9885 

9957 

••29 

•lOI 

•i"?^ 

•245 

72 

6»S 

780317 

0389 

0461 

o533 

o6o5 

0677 

0749 

0821 

0893 

0965 

72 

604 

,  1037 

1 109 

1181 

1253 

i324 

1396 

1468 

1 540 

1612 

1684 

72 

605 

1755 

1827 

1899 

1971 

2042 

2114 

2186 

2  258 

2329 

2401 

72 

606 

2473 

2544 

2616 

2688 

2759 

283 1 

2902 

2974 

3o46 

3ii7 

72 

607 

3189 

3260 

3332 

3403 

3473 

3546 

36i8 

3689 

44o3 

3761 

3832 

71 

608 

3904 

3975 

4046 

4118 

4189 

4261 

4332 

4475 

4546 

71 

609 

4617 

4689 

4760 

483 1 

4902 

4974 

5o45 

5ii6 

5187 

5259 

V 

610 

785330 

5401 

5472 

5543 

56i5 

5686 

5757 

5828 

5899 

5970 

V 

611 

6041 

6112 

6i83 

6254 

6325 

6396 

6467 

6538 

6609 

6680 

71 

612 

6751 

6822 

6893 

6964 

7035 

7106 

7177 

7248 

7319 

7390 

71 

618 

7460 
8168 

7531 

7602 

7673 

7744 

7815 

7885 

7936 

8027 

8098 

71 

614 

8239 

83io 

838 1 

845i 

8522 

8593 

8663 

8734 

8804 

71 

615 

8875 

8946 

9016 

9087 

9157 

9228 

9299 

9369 

9440 

9510 

71 

616 

9381 

9601 

9722 

9792 

9863 

9933 

•••4 

••74 

•144 

•2l5 

70 

617 

790285 

0356 

0426 

0496 

o567 

ob37 

0707 

0778 

0848 

0918 

70 

618 

0988 

1039 

1129 

1 199 

1269 

1 340 

1410 

1480 

i55o 

1620 

70 

619 

1691 

1761 

i83i 

1901 

1971 

2041 

2111 

2181 

2252 

2322 

70 

620 

792392 

2462 

2532 

2602 

2672 

2742 

2812 

2882 

2952 

3022 

70 

621 

3092 

3i62 

323i 

33oi 

3371 

3441 

35ii 

358i 

365i 

3721 

70 

622 

'    3790 

386o 

3930 

4000 

4070 

4.39 

4209 

4279 

4349 

4418 

70 

623 

4488 

4558 

4627 

4697 

4767 

4836 

4906 

4976 

5045 

5ii5 

70 

624 

;    5l85 

5254 

5324 

5393 

5463 

5532 

5602 

5672 

5741 

58ii 

70 

625 

1   5880 

5949 

6019 
67.3 

6088 

6i58 

6227 

6297 

6366 

6436 

65o5 

69 

626 

1   6574 

6644 

6782 

6852 

692. 

6990 

7060 

7129 

7198 

^ 

627 

1   726a 

7337 

7406 

7475 

7545 

7614 

7683 

7732 

7821 

& 

69 

6'i8 

It, 

8029 

8098 

8167 
8858 

8236 

83o5 

8374 

8443 

85i3 

69 

629 

8720 

8789 

8927 

8996 

9065 

9«34 

9203 

9272 

69 

630 

i  799341 

9409 

9478 

9547 

9616 

9685 

9754 

9823 

9892 

9961 

69 

631 

800029 

0098 

0167 

0236 

o3o5 

0373 

0442 

o5ii 

o58o 

0648 

69 

632 

1   0717 

0786 

o854 

0923 

0992 

1061 

1129 

1 198 

1266 

i335 

69 

633 

1404 

1472 

1 541 

1609 

1678 

1747 

I8l3 

1884 

1952 

2021 

69 

634 

2089 

2i58 

2226 

2295 

2363 

2432 

25oo 

2568 

2637 

2705 

69 

635 

2774 
3457 

2842 

2910 

2979 

3o47 

3ii6 

3i84 

3252 

3321 

3389 

68 

636 

3525 

3?94 

3662 

3730 

3798 

3867 

3935 

4oo3 

4071 

68 

637 

4139 

4208 

4276 

4344 

4412 

4480 

4548 

4616 

4685 

4753 

68 

638 

4821 

4889 

4957 

5o25 

5093 

5i6i 

5229 

5297 

5365 

5433 

68 

639 

55oi 

5569 

5^37 

5705 

5773 

5841 

5oo8 

5976 

6044 

6112 

68 

K. 

1    " 

1 

2 

8 

4 

5 

6 

7' 

8 

9 

D. 

LOGARITHMS   OF   NUMBERS. 


21 


N. 

0 

1 

2 

8 

4 

5 

6 

7 

8 

9 

D. 

640 

806180 

6248 

63i6 

6384 

645 1 

65i9 

6587 

6655 

6723 

6790 

68 

641 

6858 

6926 

6994 

7061 

7129 

7197 

7264 

7332 

7400 

l^^J 

68 

642 

7535 
8211 

7603 

7670 
8346 

7738 

7806 

7873 

7941 

8008 

8076 

8143 

68 

643 

ll'S 

8414 

8481 

8549 

8616 

8684 

8751 

8818 

67 

644 

8886 

9021 

9088 

9i56 

9223 

9290 

9358 

9425 

9492 

67 

645 

9560 

9627 

9694 

9762 

9829 

9896 

9964 

••3 1 

••98 

•i65 

67 

646 

810233 

o3oo 

o367 

0434 

o5oi 

0369 

0636 

0703 

0770 

0837 

6n 

647 

0904 

0971 

1039 

1106 

11-3 

1240 

i3o7 

1 374 

1441 

i5o8 

67 

648 

1575 

1642 

1709 

1776 

1843 

1910 

1977 

2044 

2111 

2178 

67 

649 

2245 

23j2 

2379 

2445 

25l2 

2379 

2646 

2713 

2780 

2847 

67 

650 

812913 

2980 

3o47 

3ii4 

3i8i 

3247 

33i4 

338i 

3448 

35i4 

67 

651 

3581 

3648 

3714 

3781 

3848 

3914 
438i 

3981 

4048 

4114 

4181 

67 

652 

4248 

43i4 

438i 

4447 

45i4 

4647 

4714 

4780 

4847 

67 

653 

4913 

4980 

5046 

5ii3 

5i79 

5246 

53 1 2 

5378 

5445 

55ii 

65 

654 

5578 

5644 

5711 

5777 

5843 

5910 

5976 

6042 

6109 

6175 

66 

655 

6241 

63o8 

6374 

6440 

65o6 

6573 

6639 

6705 

6771 

6838 

66 

£56 

6904 

6970 

7o36 

7102 

7169 

7235 

7301 

7367 

7433 

7499 
8160 

66 

657 

7565 

763 1 

U 

7764 

7830 

7896 

7962 

8028 

8094 

66 

658 

8226 

8292 

8424 

8490 

8556 

8622 

8688 

8754 

882c 

66 

659 

8885 

8951 

9017 

9083 

9149 

9215 

9281 

9346 

9412 

9478 

66 

6G0 

8t9544 

9610 

9676 

9741 

9807 

9873 

9939 

•••4 

••70 

•i36 

66 

661 

820201 

0267 

0333 

0399 

0464 

o53o 

0095 

0661 

0727 

0792 

66 

662 

o858 

0924 

0989 

io55 

1120 

1186 

I25l 

i3i7 

i3S2 

1448 

66 

6G3 

i5i4 

i579 

1645 

1710 

1775 

1841 

1906 

1972 

2037 

2io3 

65 

664 

2168 

2233 

2299 
2952 

2364 

243o 

2495 

256o 

2626 

2691 

2756 

65 

GG5 

2822 

2887 

3oi8 

3o83 

3i48 

32i3 

3279 

3344 

3409 

65 

666 

3474 

3539 

36o5 

3670 

3735 

3Soo 

3865 

3930 

3996 

4061 

65 

667 

4126 

4191 

4256 

4321 

43% 

445i 

45i6 

458i 

4646 

471 1 

65 

663 

4776 

4841 

4906 

4971 

5o36 

5ioi 

5i66 

523i 

5296 

536i 

65 

669 

5426 

5491 

5556 

5621 

5686 

5751 

58i5 

588o 

5945 

6010 

65 

670 

826075 

6140 

6204 

6269 

6334 

6399 

6464 

6528 

6593 

6658 

65 

671 

6723 

6787 

6852 

6917 

6981 

7046 

7111 

7175 

7240 

73o5 

65 

672 

7369 

7434 
8080 

7499 

7563 

7628 

7692 

7757 

7821 

7886 

795i 

65 

678 

8oi5 

8144 

8209 

8273 

8338 

8402 

8467 

853 1 

8095 
9239 

64 

674 

8660 

8724 

8789 

8S53 

8918 
9561 

8982 

9046 

91U 

9175 

64 

675 

9304 

9368 

9432 

9497 

9625 

9690 

9754 

9818 

9882 

64 

676 

9947 

••11 

••75 

•i39 

•204 

•268 

•332 

•396 

•460 

•525 

64 

677 

83o589 

0653 

0717 
1358 

0781 

0845 

0909 

0973 

1037 

1102 

1 166 

64 

678 

I230 

1294 
1934 

1422 

i486 

i55o 

1614 

1678 

1742 

1806 

64 

679 

1870 

1998 

2062 

2126 

2189 

2253 

23i7 

23Si 

2445 

64 

680 

832509 

2573 

2637 

2700 

2764 

2828 

2892 

2956 

3020 

3o83 

64 

6S1 

3i47 

32II 

3275 

3338 

3402 

3466 

353o 

3593 

3657 

3721 

64 

682 

3784 

3848 

3912 

3975 

4039 

4io3 

4166 

423o 

4294 

4357 

64 

6S8 

4421 

4484 

4548 

461 1 

4675 

4739 

4802 

4866 

4929 

4993 

64 

684 

5o56 

5l20 

5i83 

5247 

53io 

5373 

5437 

55oo 

5564 

5627 

63 

685 

5691 

5754 

5817 

588 1 

5944 

6007 

6071 

6i34 

6197 

6261 

63 

686 

6324 

6387 

645 1 

65i4 

6077 

6641 

6704 

6767 

6-830 

6S94 

63 

687 

6957 

7020 

70S3 

7146 

7210 

7273 

7336 

7399 

7462 

7325 

81 56 

63 

688 

■7388 

7652 
8282 

77i5 

7778 

7841 

7904 

It] 

8o3o 

8093 

63 

689 

8219 

8345 

8408 

8471 

8534 

8660 

8723 

8786 

63 

690 

838849 

8912 

8975 

9o38 

9101 

9«64 

9227 

9289 

9352 

94i5 

63 

691 

9478 

9541 

9604 

9667 

9729 

9792 

9855 

9918 

9981 

••43 

63 

692 

840106 

0169 

0232 

0294 

0357 

0420 

04S2 

o545 

0608 

0671 

63 

693 

0733 

0796 

o859 

0921 

0984 

1046 

1 109 

1172 

1234 

1297 

63 

694 

1359 

1422 

1485 

1 547 

1610 

1672 

1735 

1797 

i860 

1922 

63 

695 

19S5 

2047 

2110 

2172 

2235 

2297 

2360 

2422 

2484 

2047 

62 

696 

2609 

2672 

2734 

2796 

2859 

2921 

2983 

3046 

3io8 

3170 

62 

697 

.3233 

3295 

3357 

3420 

348? 

3544 

36o6 

.3669 

3731 

3793 

62 

698 

3855 

3918 

3980 

4042 

4104 

4166 

4229 

4291 

4353 

44i5 

62 

699 

4477 

4539 

4601 

4664 

4726 

4788 

4b5o 

4912 

4974 

5o36 

62 

N. 

0 

1 

2 

8 

4 

5 

6 

7 

8 

9 

D. 

.-)9 


LOGARITHMS    OF   XOIBERS. 


K. 

:   0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

700 

845098  5i6o 

5222 

5284 

5346 

5408 

5470 

5532  j  5594 

5656 

62 

701 

5718  5780 

5842 

5904 

5966 

6028 

6090 

6i5i  1  62i3 

6275 

62 

702 

6337  6399  6461 

6323 

6585 

6646 

6708 

6770  6832 

6894 

62 

703 

6955  7017  7079 

7141 

7202 

7264 

7326 

7388 

^ 

7511 
8128 

62 

704 

:   7373  7634 

8189   8231 

;   880!  8866 

7696 

7758 

7819 

7881 

7943 

8004 

62 

705 

83.2 

8374 

8433 

8497 

8559 

8620 

8682 

8743 

62 

706 

8928 

8989 
9604 

905 1 

9112 

9174 

9235 

9297 

9358 

61 

707 

9419  9481 

9542 

9663 

9726 

9788 

9849 

9911 

9972 

61 

70S 

85oo33  0095 

0156 

0217 

0279 

o34o 

0401 

0462 

0324 

o585 

6t 

7oy 

i   0646  0707 

0769 

o83o 

0S91 

0952 

1014 

1075 

1x36 

1197 

61 

710 

;  851258 

l320 

i38i 

1442 

i5oJ 

1 564 

1625 

1686 

1747 

1809 

61 

711 

1870 

1931 

1992 

2o53 

2114 

2175 

2236 

2297 

2358 

2419 

61 

712 

2480 

2341 

2602 

2663 

2724 

2785 

2846 

2907 

2968 

3029 

61 

71S 

1   3090 

3i5o 

3211 

3272 

3333 

3394 

3455 

35i6 

3577 

3637 

61 

714 

3698 

3759 

3820 

388i 

3941 

4002 

4o63 

4124 

4i85 

4245 

61 

715 

i   43o6 

4367 

4428 

4488 

4549 

4610 

4670 

4731 

4792 

4852 

61 

716 

1   4913 

4974 

5o34 

5095 

5i56 

5216 

5277 

5337 

5398 

5459 

61 

717 

5519 

5d8o 

5640 

5701 

5761 

5822 

5882 

5943 

6oo3 

6064 

61 

71S 

6124 

6i85 

6245 

63o6 

6366 

6427 

6487 

6348 

6608 

6668 

60 

719 

'   6729 

6789 

685o 

6910 

6970 

703 1 

7091 

7i52 

7212 

7272 

60 

720 

1  857332 

7393 

7453 

75i3 
8116 

7574 

7634 

7694 

7755 

7815 

7875 

60 

721 

7935 
8537 
9i38 

7995 

8o56 

8176 

8236 

8297 

8357 

8417 

8477 

60 

722 

8597 

8657 

8718 

8778 

8838 

8898 

8958 
9359 
•i58 

9018 

9078 

60 

723 

9198 

9^?^ 

9318 

9379 

9978 

9439 

9499 
••98 

9619 

9679 

60 

724 

selhl 

ti^ 

9839 

9918 

••33 

•218 

•278 

60 

725 

0458 

o5.8 

0578 

0637 

0697 

0757 

0817 

0877 

60 

726 

1   0937 

X 

1036 

1116 

1176 

1236 

1295 

i355 

I4i5 

1475 

60 

727 

;   1534 

1 654 

1714 

1773 

1 833 

1893 
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1952 

2012 

2072 

60 

728 

i    2l3l 

2I9I 

2787 

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23l0 

2370 

243o 

2549 

260S 

2668 

60 

729 

I   2728 

2847 

2906 

2966 

3o25 

3o85 

3i44 

3204 

3263 

60 

730 

:  863323 

3382 

3442 

35oi 

356i 

3620 

368o 

3739 

3799 

3858 

59 

731 

i   3917 

3977 

4o36 

4689 

4155 

4214 

4274 

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4392 

4452 

59 

732 

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4370 

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4748 

4808 

4867 

4926 

4985 

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733 

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5i63 

5222 

5282 

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5400 

5459 

55i9 

5578 

5637 

59 

734 

i   6287 

5755 

58i4 

5874 

5933 

5992 

605T 

6110 

6169 

6228 

59 

735 

6346 

6405 

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6583 

6642 

6701 

6760 

6819 

59 

736 

1   6878 

6937 

6996 

7055 

7114 

7173 

7232 

8468 

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7409 

59 

737 

7467 

7526 

7585 

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7762 

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1   8o56 

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739 

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8821 

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59 

740 

869232 

9290 

9349 

9408 

9466 

9525 

9584 

9642 

9701 

9760 

^ 

741 

9818 

9877 

9933 

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ll 

742 

870404 

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n]t 

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1047 

1106  j  II64 

1223 

1281 

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1398 

1456 

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744 

!   1573 
1   2i56 

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1690  1748 

1806 

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1923 

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2040 

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58 

745 

22l5 

2273 

233i 

2389 

2448 

2306 

2564 

2622 

58 

746 

'   2739 

2797 

2855 

2913 

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3o3o 

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3204 

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747 

:   3321 

3379 

3437 

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36ii 

3669 

3727  1  3785 

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58 

743 

1   3902 

3960 

4018 

4076 

4i34 

4192 

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4366 

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749 

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4540 

4598 

4656 

4714 

4772 

4830 

4888 

4945 

5oo3 

58 

750 

875061 

5119 

5i77 

5235 

5293 

535i 

5409 

5466 

5524 

5582 

58 

751 

5640 

5698 

5756 

58i3 

5871 

5929 

5087 

6045 

6102 

6160 

58 

752 

6218 

6276 

6333 

6391 

6449 

6564 

6622 

6680 

6737 

58 

753 

6795 

6853 

6910 

6968 

7026 

7083 

7141 

7199  !  7256 

73.4 

58 

754 

7371 

7429 

7487 

7544 

7602 

7659 

7717 

7-774 

7832 

7889 

58 

755 

1;   7947 

8004 

8062 

81 19 

8177 

8234 

8292 

8349 

8407 

8464 

57 

756 

;   8522 

8579 

8637 

8694 

8752^ 

8809 

8866 

8924 

8981 

9039 

57 

757 

1   9096 

9.53 

9211 

9268 

9325^ 

9383 

9440 

9497 

9555 

9612 

57 

753 

•1   9669 

9726 

9-84 

9>^4i  i  9^98 

9936 

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••70 

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57 

759 

,|  880242 

0299 

o356 

o4i3  j  0471 

o528 

o585 

0642 

0699 

0756 

57 

N. 

f   0 

1 

2 

3     4 

5 

6  1  7   1  8 

9 

D. 

LOGARITHMS   OF   NUMBERS. 


23 


N. 

1   0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1- 

760 

880814 

0871 

0928 

0985 

1042 

1099 

ii56 

12l3 

1271 

i328 

57 

7G1 

1385 

1442 

1499 

1 556 

i6i3 

1670 

1727 

1784 

1841 

1898 

57 

762 

1955 

2012 

2069 

2126 

2i83 

2240 

2297 

2354 

2411 

2468 

57 

763 

2525 

258i 

2638 

2695 

2752 

2809 

2866 

2923 

2980 

3o37 

57 

764 

3093 

3i5o 

3207 

3264 

3321 

3377 

3434 

3491 

3548 

36o5 

57 

765 

366i 

3718 

3775 

3832 

3888 

3945 

4002 

4059 

4n5 

4172 

57 

766 

4229 

4285 

4342 

4399 
4965 

4455 

4512 

^5^ 

4625 

4682 

4739 

57 

767 

4795 

4852 

4909 

5022 

5078 

5192 

5248 

53o5 

5i 

76S 

536i 

5418 

5474 

553 1 

5587 

5644 

5700 

5757 

5di3 

58-0 

57 

769 

5926 

5983 

6039 

6096 

6i52 

6209 

626D 

6321 

6378 

6434 

56 

770 

886491 

6547 

6604 

6660 

6716 

6773 

6829 

6885 

8067 

6998 

56 

771 

7054 

71U 

7167 

7223 

7280 

7336 

7392 

79D3 

7449 

7561 

56 

772 

7617 

7674 

8292 

7786 

7842 

7898 

8011 

8123 

56 

773 

8179 

8236 

8348 

8404 

8460 

85i6 

8573 

8629 

8685 

56 

774 

8741 

879T 

8853 

8909 

8965 

9021 

9077 

9134 

9190 

9246 

56 

775 

9302 

9358 

9414 

9470 

9526 

9582 

9638 

9694 

9750 

9806 

56 

776 

9862 

9918 

9974 

••3o 

••86 

••141 

•197 

•253 

•309 

•365 

56 

777 

890421 

0477 

0333 

0589 

0645 

0700 

07  D6 

0812 

0868 

0924 

56 

778 

^% 

io35 

1091 

1147 

1203 

1259 

i3i4 

1370 

1426 

1482 

56 

779 

1593 

1649 

1705 

1760 

1816 

1872 

1928 

1983 

2039 

56 

780 

892095 

2l50 

2206 

2262 

23i7 

2373 

2429 

2484 

2540 

2595 

56 

781 

265i 

2707 

2762 

2818 

2873 

2929 

298D 

3o4o 

3096 

3i5i 

56 

782 

8207 

3262 

33i8 

3373 

3429 

3484 

3D40 

3595 

3651 

3706 

56 

783 

3762 

38i7 

3873 

3928 

3984 

4039 

4094 

4i5o 

42o5 

4261 

55 

.784 

43i6 

4371 

4427 

4482 

4538 

4593 

4648 

4704 

4739 

4814 

55 

785 

4870 

4925 

4980 

5o36 

5091 

5i46 

5201 

5257 

53i2 

5367 

55 

786 

5423 

5478 

5533 

5588 

5644 

5699 

5754 

5809 

5864 

5920 

55 

787 

5975 
6526 

6o3o 

6o85 

6140 

6195 

6251 

63o6 

6361 

6416 

6471 

55 

788 

658i 

6636 

6692 

6747 

6802 

6857 

6912 

6967 

7022 

55 

78y 

7077 

7132 

7187 

7242 

7297 

7352 

7407 

7462 

7517 

7572 

55 

790 

897627 

7682 

7737 

llf, 

7847 

7902 
8451 

79^7 

8012 

8067 

8122 

55 

791 

8176 

823i 

8286 

8396 

8D06 

856i 

861 5 

8670 

55 

792 

8725 

8780 

8835 

8890 
9437 

8944 

8999 

9054 

9109 

9164 

9218 

55 

793 

9273 

9328 

9383 

9492 

9347 

9602 

9656 

9711 

9766 

55 

794 

9821 

9875 

9930 

9985 

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••94 

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55 

795 

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0476 

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0640 

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0859 

55 

796 

0913 

0968 

1022 

1077 

ii3i 

1186 

1240 

1295 

1 349 

1404 

55 

797 

1458 

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1567 

1622 

1676 

1731 

1785 

1840 

1894 

1948 

54 

798 

2003 

2057 

2112 

2166 

222Z 

2275 

2329 

2384 

2438 

2492 

54 

799 

2547 

2601 

2655 

2710 

2764 

2818 

2873 

2927 

2981 

3o36 

54 

800 

903090 

3c44 

3199 

3253 

3307 

336i 

3416 

3470 

3524 

3578 

54 

801 

3633 

3687 

3741 

3795 

3849 

3904 

3958 

4012 

4066 

4120 

54 

802 

4174 

4229 

4283 

4337 
4878 

4391 

4445 

4499 

4553 

4607 

466i 

54 

803 

4716 

4770 

4824 

4932 

49S6 

5040 

5094 

5 1 48 

5202 

54 

804 

5256 

53io 

5364 

5418 

5472 

5526 

558o 

5634 

5688 

5742 

54 

805 

5796 

585o 

5904 

5958 

6012 

6066 

6119 
6658 

6173 

6227 

6281 

54 

806 

6335 

6389 

6443 

6497 

655i 

6604 

6712 

6766 

6820 

54 

807 

6874 

6927 

6981 

7035 

7089 

7143 

7196 

725o 

7304 

7358 

54 

808 

741 1 

7465 

7519 
8o56 

7573 

7626 

7680 
8217 

7734 
8270 

7787 

7841 
8378 

tt 

54 

809 

7949 

8002 

8110 

8i63 

8324 

54 

810 

908485 

8539 

8592 

8646 

8699 

8753 

8807 

8860 

8914 

8967 

54 

811 

9?L' 

9074 

9128 

9181 

9235 

9289 

9342 

9396 

9449 

95o3 

54 

812 

9556 

9610 

9663 

9716 

9770 

9823 

9»77 

9930 

9984 

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53 

813 

910091 

0144 

0107 
0731 

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o358 

0411 

0464 

o5i8 

0571 

53 

814 

0624 

0678 

0784 

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0891 

0944 

OQ98 

io5i 

1104 

53 

815 

ii58 

1211 

1264 

i3i7 

1371 

1424 

1477 

i53o 

i584 

1637 

53 

816 

i6go 

1743 

1797 

i85o 

1903 

1956 

2009 

2063 

2116 

2169 

53 

817 

2222 

2275 

2328 

238i 

2435 

2488 

2541 

2594 

2647 

2700 

53 

818 

2753 

2806 

2859 

2913 

2966 

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3178 

3^31 

53 

819 

3284 

3337 

3390 

3443 

3496 

3549 

36o2 

3655 

3708 

3761 

53 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

24 


LOGAEITHMS   OF  NUMEEKS. 


N. 
8'20 

1   ^ 

1 

2 

8 

4  • 

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f6l 

■  7  - 

:  8 

9 

D. 

■  9i38i4 

3867 

3920 

45o2 

4026 

4070  ~ 
4608 

4i32 

4184 

4237 

4290 

53 

8:^1 

4343 

4396 

4449 

4555 

4660 

4713 

4766 

4819 

53 

8'J2 

4872 

4925  1  4977 

5o3o 

5o83 

5i36 

5189 

5241 

5294 

5347 

.53 

82S 

i   5400 

5453 

DD05 

5558 

56ii 

,5664 

5716 

5769 

5822 

5875 

53 

8'J-l 

1   59-^7 

5980 
65o7 

6o33 

6o85 

6i38 

6191 

6243 

6296 

6340 
6875 

6401 

53 

8'ir, 

6454 

6559 

6612 

6664 

6717 
7243 

6770 

6822 

6927 

53 

82(1 

6980 
75o6 

7033 

7085 

7.38 

7190 

7295 

7348 

7400 

7433 

53 

827 

7558 
8o83 

7611 

7663 

7716 

7768 

7820 

7873 

7925 

SS 

52 

82S 

8o3o 

8i35 

8188 

8240 

8293 

8345 

8397 

845o 

52 

829 

8555 

8607 

8659 

8712 

8764 

88i6' 

8869 

89,21 

8973 

9026 

52 

830 

;  919078 

9i3o  9183 

9235 

9287 

9340 

9392 

9444 

9496 

9549 

52 

831 

9601 

9653  9706 

9758 

9810 

9862 

9914 

9967 

••.9 

••71 

52 

832 

■  920123 

0176 

0228 

0280 

o332 

o384 

0436 

0489 

o54i 

0593 

52 

S33 

i   0645 

0697 

0749 

0801 

0853 

0906 

0958 

lOIO 

1062 

1114 

52 

834 

;   1 166 

1218 

1270 

l322 

1374 

1426 

1478 

i53o 

1 582 

1 634 

52 

835 

i   1686 

1738 

1790 

1842 

1894 

1946 

2??8 

2o5o 

2102 

2.54 

52 

836 

2206 

2258 

23lO 

2362 

2414 

2466 

2570 

2622 

2674 

52  , 

837 

2725 

2777 

2829 
3348 

2881 

2933 

2985 

3o37 

3089 

3i4o 

3.92 

52 
52 

838 

3244 

3296 

3399 

345i 

35o3 

3555 

3607 

3658 

3710 

830 

3762 

38i4 

3865 

3917 

3969 

4021 

4072 

4124 

4176 

4228 

52 

840 

924279 

4331 

4383 

4434 

4486 

4538 

4589 

4641 

4693 

4744 

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841 

4796 

4848 

4899 

4951 

5oo3 

5o54 

5io6 

5i57 

52og 

5261 

52 

S42 

53i2 

5364 

541 5 

5467 

55i8 

5570 

5621 

5673 

5725 

5776 

52 

843 

5828 

5879 

5931 

5982 

6o34 

6o85 

6137 

6188 

6240 

6291 

5i 

844 

6342 

6394 

6445 

6497 

6548 

6600 

665 1 

6702 

6754 

68o5 

5i 

845 

6857 

6908 

6959 

701 1 

7062 

7114 

7i65 

7216 

7268 

7319 

5i 

846 

7370 

7422 

7473 

8037 

7576 

7627 

7678 
8191 

7730 

7781 
8293 

7832 
8345 

5i 

847 

i   7883 

7935 

7986 

8088 

8140 

8242 

5i 

848 

1   8396 

8447 

8498 

8549 

8601 

8652 

8703 

8754 

88o5 

8857 
9368 

5i 

849 

1   8908 

8959 

9010 

9061 

9112 

9163 

9215 

9266 

9317 

5i 

850 

929419 

9470 

9521 

9572 

9623 

tii 

9725 

9776 

9827 

9579 

5i 

851 

I   9930 

9981 

••32 

••83 

•i34 

•236 

•287 

•338 

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0898 

5i 

852 

j  930440 

0491 

o542 

0592 

0643 

0694 

0745 

0796 

0847 

5i 

853 

'   0949 

1000 

io5i 

1102 

ii53 

1204 

1254 

i3o5 

i355 

1407 

5i 

854 

i   1458 

1 509 

i56o 

1610 

1661 

1712 

1763 

1814 

1 865 

1915 

5i 

855 

S   1966 

2017 

2068 

2118 

2169 

2220 

2271 

2322 

2372 

2423 

5i 

856 

2474 

2524 

2575 

2626 

2677 

2727 

2778 

2S29 

ll^ 

2930 

5. 

857 

29S1 

3o3i 

3082 

3i33 

3i83 

3234 

3285 

3335 

3437 

5i 

858 

3487 

3538 

3589 

3639 

3690 

3740 

3791 

3841 

3892 

3943 

5i 

859 

3993 

4044 

4094 

4145 

4195 

4246 

4296 

4347 

4397 

4448 

5i 

860 

:  934498 

4549 

4599 

465o 

4^00 

475i 

4801 

4852 

4902 

4953 

5o 

861 

1   5oo3 

5o54 

5io4 

5i54 

5203 

5255 

53o6 

5356 

5406 

5457 

5o 

862 

!   5507 

5558 

56o8 

5658 

5709 

5759 

5809 

5860 

5910 

5960 

5o 

8'13 

601 1 

6061 

6111 

6162 

6212 

6262 

63 1 3 

6363 

64.3 

6463 

5o 

864 

65i4 

6564 

6614 

6665 

6715 

6765 

68i5 

6865 

6916 

6966 

5o 

865 

7016 

7066 

7117 

7167 

7217 
7718 

'7267 

73.7 

7367 

74.8 

7468 

5o 

866 

75i8 

7568 

7618 

7668 

7769 
8269 

7819 

7869 

7919 

7969 

5o 

867 

8019 

8069 

&119 

8169 

8219 

8320 

8370 

8420 

8470 

5o 

868 

8520 

8570 

8620 

8670 

8720 

8770 

8820 

8870 

8920 

8970 

5o 

869 

9020 

9070 

9120 

9170 

9220 

9270 

9320 

9369 

9419 

9469 

5o 

870 

939519 

9569 
0068 

9610 
0118 

9669 

9719 
0218 

9769 

9819 

9869 

99.8 

9968 

5o 

871 

940018 

0168 

0267 

o3i7 

0367 

0417 

0467 

5o 

872 

o5i6 

o566 

c6i6  0666  1 

0716 

0765 

081 5 

0865 

0915 

0964 

5o 

873 

1014 

1064 

1114 

Ji63 

12-. 3 

1263 

i3i3 

i362 

14.2 

1462 

5o 

874 

j5ii 

i56i 

1611 

1660 

1710 

1760 

1809 

1809 

1909 

.958 

5o 

875 

2008 

2o58 

2107 

2137 

2207 

2256 

23o6 

235D 

240D 

2455 

5o 

876 

25o4 

2554 

2603 

2653 

2702 

2752 

2801 

285 1 

2901 

2950 

5o 

877 

3ooo 

3o49 

3099 

3148 

3.98 

3247 

3297 

3346 

3396 

3445 

49 

878 

3495 

3544 

3593 

3643 

36q2 

3742 

3791 

3841 

3890 

3939 

49 

879 

3989 

4o38 

4088 

4i37 

4186 

4236 

4285 

4335 

4384 

•4433 

49 

N. 

0      1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

LOGARITHMS   OF   NUMBERS 


25 


N. 

' 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

880 

1  944483 

4532 

458i 

463 1 

4680 

4729 

4779 

4828  4877  1  4927 

49 

8S1 

4976 

5o25 

5074 

5i24 

5173 

5222 

5272 

5321  5370  ;  5419 

49 

882 

5469 

55i8 

5567 

56i6 

5665 

5715 

5764 

58 1 3  5862 

5912 

49 

883 

5961 

6010 

6059 

6108 

6i57 

6207 
6698 

6256 

63o5 

6354 

6403 

49 

884 

6402 

65oi 

655i 

6600 

6649 

6747 

6796 

6845 

6894 

49 

885 

6943 

6992 

7041 

7090 

7140 

7189 

7238 

7287 

7336 

7385 

49 

886 

7434 

7453 

p32 
8022 

^07^ 

7630 
8119 

7679 
8168 

7728 

7777 
8266 

7826 
83i5 

7875 
8364 

49 

887 

7924 
8413 

ml 

8217 

49 

883 

85ii 

8560 

8609 

8657 

8706 

8755 

8804 

8853 

49 

889 
890 

8902 

8951 

8999 

9048 

9097 

9146 

9195 

9244 

9292 

9341 

49 

949390 

9439 

9488 

9536 

9585 

9634 

9683 

9731 

9780 

9829 

49 

891 

9878 

9926 

9975 

••24 

••73 

•J21 

•170 

•219 

•267  ^316 

49 

892 

95o365 

0414 

0462 

o5ii 

o563 

0608 

0657 
n43 

0706 

0754 

o8o3 

49 

893 

o85i 

0900 

0949 

0997 

1046 

1095 

1192 

1240 

1289 

49 

894 

i338 

1386 

1435 

i4«3 

1 532 

i58o 

1629 

1677 

1726 

1773 

S 

895 

1823 

1872 

1920 

1969 

2017 

2066 

2114 

2i63 

2211 

2260 

896 

23o8 

2356 

24o5 

2453 

2502 

255o 

2599 

2647 

2696 

2744 

48 

897 

2792 

2841 

2889 

2938 

2986 

3o34 

3o83 

3i3i 

3i8o 

3228 

48 

898 

3276 

3325 

3373 

3421 

3470 

35i8 

3566 

36i5 

3663 

3711 

48 

899 

3760 

38o8 

3856 

3905 

3953 

4001 

4049 

4098 

4146 

4194 

48 

900 

954243 

4291 

4339 

4387 

4435 

4484 

4532 

458o 

4628 

4677 

48 

901 

4725 

4773 

4821 

4S69 

4918 

4966 

5oi4 

5o62 

5iio 

5i58 

48 

902 

5207 
5688 

5255 

53o3 

535i 

5399 

5447 

5495 

5543 

5592 

5640 

48 

903 

5736 

5784 

5832 

5880 

5928 

5976 

6024 

6072 

6120 

48 

904 

6168 

6216 

6265 

63i3 

636 1 

6409 

6457 

65o5 

6553 

6601 

48 

•905 

6649 

6697 

6745 

6793 

6840 

6888 

6936 

6984 

7032 

7080 

48 

906 

7128 

7176 

7324 

7272 

7320 

7368 

7416 

7464 

75x2 

m 

48 

907 

7607 

7655 

7703 

775i 
8229 

7799 
8277 

7847 

7894 

7942 
8421 

7990 

48 

908 

8086 

8i34 

8181 

8325  8373 

8468 

85i6 

48 

i*09 

8564 

8612 

8659 

8707 

8755 

88o3 

885o 

8898 

8946 

8994 

48 

910 

959041 

^^ 

9137 

9i85 

9232 

9280 

9328 

9375 

9423 

9471 

48 

911 

9518 

9614 

9661 

9709 

9757 

9804 

9852 

9900 

9947 

48 

912 

9995 

••42 

••90 

•i38 

•i85 

•233  ^280 

•328 

•376 

•423 

48 

913 

960471 

o5i8 

o566 

o6i3 

0661 

0709  0756 

0804 

o85i 

0S99 

48 

914 

0946 

0994 

1041 

1089 

ii36 

1 184 

I23l 

1279 

i326 

1374 

47 

915 

1421 

1469 

i5i6 

i563 

1611 

1658 

1706 

1753 

1801 

1848 

47 

916 

1895 

1943 

1990 

2o38 

2o85 

2l32 

2180 

2227 

2275 

2322 

47 

917 

2369 

2843 

2417 

2404 

25ll 

2559 

2606 

2653 

2701 

2748 

2795 

47 

918 

2890 

2937 

2985 

3o32 

3079 

3126 

3174 

3221 

3268 

47 

919 

33i6 

3363 

3410 

3457 

35o4 

3552 

3599 

3646 

3693 

3741 

47 

920 

963788 

3835 

3882 

3929 

3977 
44.i8 

4024 

4071 

4118 

4i65 

4212 

47 

9-21 

4260 

4307 

4354 

4401 

4495 

4542 

4590 

4637 

4684 

47 

922 

473 1 

477« 

4825 

4-S72 

4919 

4966 

5oi3 

5o6i 

5io8 

5i55 

47 

923 

5202 

5249 

5296 

5343 

5390 

5437 

5484 

553 1 

5578 

5625 

47 

924 

5672 

5719 

5766 

58i3 

586o 

5907 

5954 

600 1 

6048 

6093 

47 

925 

6142 

6189 
6658 

6236 

6283 

6329 

6376 

6423 

6470 

65i7 

6564 

47 

926 

6611 

6705 

6752 

6799 

6845 

6892 

6939 

6986 

7033 

47 

927 

7080 

7127 

7173 

7220 

7267 

7314 

7361 

7408 

7454 

7501 

47 

928 
929 

.7548 

8oi6 

ui 

7642 
8109 

7688 
8i56 

7735 
8io3 

7782 
8249 

7829 
8296 

llll 

7922 
8390 

Itl 

47 
47 

930 

'  968483 

853o 

8576 

8623 

86-0 

8716 

8763 

8810 

8856 

8903 

47 

931 

i   8950 

8996 

9043 

9090 

9136 

9.83 

9229 

9276 

9323 

9369 

47 

932 

9416 

9463 

9009 

9556 

9602 

9649 

9693 

9742 

9789 

9S35 

47 

933 

9882 

9928 

9973 

••21 

••68 

•114 

•161 

•207 

•254 

•3oo 

47 

934 

1  970347 

0393 

0440 

04S6 

o5>3 

0579 

0626 

0672 

0710 
ii83 

0765 

46 

935 

0812 

o858 

0904 

0951 

09  V7 

1044 

1090 

1137 

1229 
1693 

46 

936 

1276 

l322 

1369 

1415 

14  I 

i5o8 

1 554 

i6ol 

1647 

46 

937 

1740 

1786 

1832 

1879 

1975 

1971 

2018 

2064 

2110 

2137 

46 

938 

2203 

2249 

2295 

2342 

23^8 

2434 

2481 

2527 

2573 

2619 

46 

939 

2666 

0 

2712 

2758 

2804 

283i 

2897 

2943 

2989 

3o35 

3082 

46 

N. 

1 

2 

3 

* 

5 

1  " 

7 

8 

9 

D. 

26 


LOGAEITHMS   OF   NUMBEES. 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

940 

973128 

3i74 

3220 

3266 

33i3 

3359 

34o5 

345i 

3497 

3543 

46 

941 

3590 
4o5i 

3636 

3682 

3728 

3774 

3820 

3866 

'4 

4SU 

3959 

4oo5 

46 

942 

4097 

4143 

4189 

4235 

4281 

4327 

4420 

4466 

46 

943 

45i2 

4558 

4604 

465o 

4606 

4742 

4788 

4880 

4926 

46 

944 

4972 

5oi8 

5o64 

5iio 

5.D6 

5202 

5248 

5294 

5340 

5386 

46 

945 

5432 

5478 

5524 

5570 

56i6 

5662 

5707 

5753 

5799 

5845 

46 

946 

5891 

6396 

5983 

6029 

6oi5 

6121 

6167 

6212 

6258 

63o4 

46 

947 

63  5o 

6442 

6488 

6533 

6579 

6625 

6671 

6717 

6763 

46 

948 

6808 

6854 

6900 

6946 

6992 

7037 

7083 

7129 

7175 

7220 

46 

949 

7266 

7312 

7358 

7403 

7449 

7495 

7541 

7586 

7632 

7678 

46 

950 

977724 
8181 

7769 
8226 

78.5 

7861 
83i7 

7906 

7952 

7998 

8043 

8089 

8 1 35 

46 

951 

8272 

8363 

8409 

8454 

85oo 

8546 

8591 

46 

952 

8637 

8683 

8728 

8774 

88i§ 

8863 

891 1 
9366 

8956 

9002 

9047 

46 

953 

9093 

9i38 

9184 

Q23o 

9273 

9321 

9412 

9457 

95o3 

46 

954 

9548 

9594 

9639 

9685 

9730 

9776 

9821 

9867 

X 

9958 

46 

955 

980003 

0049 

0094 

0140 

oi85 

023l 

0276 

o322 

0412 

45 

956 

0458 

o5o3 

0049 

0594 

0640 

0685 

0730 

0776 

0821 

0867 

45 

957 

0912 

0957 

ioo3 

1048 

1093 

1139 

1184 

1229 

1275 

l320 

45 

958 

1366 

I4U 

1456 

i5oi 

1547 

1592 

1637 

1 683 

1728 

1773 

45 

959 
960 

1819 

1864 

J  909 

1954 

2000 

2045 

2090 

2i35 

2181 

2226 

45 

982271 

23i6 

2362 

2407 

2452 

2491 

2543 

2588 

2633 

2678 

45 

961 

2723 

2769 

2814 

2859 

2904 

2949 

2994 

3o4o 

3o85 

3i3o 

45 

962 

3175 

3220 

3265 

33io 

3356 

3401 

3446 

3491 

3536 

358i 

45 

963 

3626 

3671 

3716 

3762 

3807 

3852 

3897 

3942 
4392 

3987 

4o32 

45 

964 

4077 

4122 

4167 

4212 

4257 

43o2 

4347 

4437 

4482 

45 

965 

4527 

4572 

4617 

4662 

4707 

4752 

4797 

4842 

4867 

4932 
5382 

45 

96^ 

4977 

5022 

5067 

5112 

5.57 

5202 

5247 

5292 

5337 

45 

967 

5426 

5471 

55.6 

556r 

56o6 

565 1 

5696 

5741 

5786 

583o 

45 

963 

5875 

6920 

6369 

5965 

6010 

6o55 

6100 

6144 

6189 

6234 

6279 

45 

969 

6324 

64i3 

6458 

65o3 

6548 

6593 

6637 

6682 

6727 

45 

970 

986772 

6817 

6861 

6906 

6951 

6996 

7040 

7085 

7i3o 

7175 

45 

971 

7219 

7264 

7309 

7353 

7398 

7443 

7488 

7532 

8024 

7622 

45 

972 

7666 
8ii3 

7711 

7756 
8202 

7800 

7845 

7890 

Ifsi 

lilt 

8068 

^l 

973 

8157 

8247 

8291 

8336 

8470 

85i4 

45 

974 

8559 

&604 

8648 

8693 

8737 

8782 

8826 

8871 

8916 

8960 

45 

975 

9000 

9049 

9094 

9i38 

9183 

9227 

9272 

93i6 

9361 

94o5 

45  , 

9:6 

9430 

9494 

95J9 

9583 

9628 

9672 

9717 

9761 

9806 

9856 

44 

t)77 

,8o5 

9903 3o 

0783 

9983 

••28 

••72 

•.17 

.•161 

•206 

•25o 

•294 
0738 

44 

978 

0428 

0472 

o5i6 

o56i 

o6o5 

o65o 

0694 

44 

979 

0827 

0871 

0916 

0960 

1004 

1049 

1093 

1137 

1182 

44 

980 

991226 

1270 

i3i5 

1359 

l4o3 

1448 

1492 

1536 

i58o 

1625 

44 

981 

^^669 

1713 

1758 

1802 

1846 

itr, 

1935 
2377 

1979 

2023 

2067 

44 

982 

2111 

2i56 

2200 

2244 

2288 

2421 

2465 

2509 

44 

983 

2554 

2598 

2642 

2686 

2730 

2774 

2819 

2863 

2907 

2951 

44 

984 

2995 

3o39 

3o83 

3127 
3568 

3172 

3216 

3260 

33o4 

3348 

33?2 

44 

985 

3436 

3480 

3524 

36i3 

3657 

3701 

3745 

3789 

3833 

44 

986 

3877 

3921 
436i 

3965 

4009 

4o53 

4097 

4141 

4i85 

4229 

4273 

44 

987 

43.7 

44o5 

4449 

4493 

4537 

458i 

4625 

4669 
5io8 

4713 

44 

9SS 

4737 

4801 

4845 

4889 

4933 

4977 

502I 

5o65 

5i52 

44 

989 

5196 

5240 

5284 

5328 

5372 

5416 

5460 

55o4 

5547 

5591 

44 

900 

995635 

5679 

5723 

5767 

58ii 

5854 

5898 

5942 

5986 

6o3o 

44 

901 

6074 

6.17 
6555 

6161 

6205 

6249 

6293  6337* 

6380 

6424 

6468 

44 

992 

65i2 

6599  6643 

6687 

6731 

6774 

6818 

6862 

6906 
7343 

44 

993 

6949 

6993 

7037 

7080 

7124 

7168 

7212 

7255 

7299 

44 

994 

7386 

7430 

7474 

7517 

7561 

7605 

7648 
8o85 

7692 

7736 
8172 

7779 
8216 

44 

995 

7823 
8259 

7867 

7910 

l%t 

7998 

8041 

8129 

44 

996 

83o3 

8347 

8434 

8477 

8521 

8564 

8608 

8652 

44 

997 

8695 

8739 

8782 

8S26 

8869 

8913 
9348 

8956 

9000 

9043 

9087 

44 

908 

9.31 

9174 

9218 

9261 

93o5 

9392 

9435 

9479 

9522 

44 

999 

9565  9609 

9652  9696 

9739 

9783 

9826 

9870 

9913 

9957 

43 

K 

0 

1 

2   1   3 

4 

5 

6 

7 

8 

9 

D. 

LOGARITHMS     OF    NUMBERS. 


27 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

1000 

000000 

0043 

0087 

oi3o 

0174 

0317 

0260 

o3o4  o347 

0391 

43 

lOOI 

0434 

0477 

052I 

o564 

0608 

o65i 

0694 

0738  i  0781 

0824 

43 

1002 

0868 

"^l 

0954 

1 388 

0998 

io4i 

iob4 

1128 

1171 1  1214 

1238 

43 

[oo3 

i3oi 

I43i 

1474 

i5i7 

I36l 

1604  1647 

1690 

43 

1004 

1734 

nil 

1820 

i863 

IK 

1950 

1993 

2o36  2080 

2123 

43 

ioo5 

002166 

2209 

2252 

2296 

2382 

2420 

2468,  25 1 2 

2.05 

43 

ioo6 

2598 

2641 

2684 

2727 

2771 

2814 

2857 

2900  2943 

2986 

43 

1007 

3029 

3073 

3ii6 

3i59 

3202 

3245 

3288 

333i 

3374 

3417 

3848 

43 

1008 

346 1 

35o4 

3547 

3390 

3633 

3676 

3719 

3762 

38o5 

43 

1009  , 

3891 

3934 

3977 

4020 

4o63 

4106 

4149 

4192 

4235 

4278 

43 

lOIO 

004321 

4364 

4407 

445o 

4493 

4536 

4579 

4622 

4665 

4708 

43 

lOII 

4751 

4794 

4837 

4880 

4923 

4966 
5395 

5009 

5438 

5o52 

5095 

5i38 

43 

I0I2 

5i8i 

5223 

5266 

5309 

5352 

5481 

5524 

5567 

43 

ioi3 

5609 

5652 

5695 

5738 

5781 

5824 

5867 

5909 

6?8o 

5995 

43 

IOI4 

6o38 

6081 

6124 

6166 

6209 

6252 

6295 

6338 

6433 

43 

lOID 

006466 

65o9 

6552 

6594 

6637 

6680 

6723 

6765 

6808 

685 1 

43 

1016 

6894 

6936 
7364 

6979 

7022 

7065 

7107 

7i5o 

7.93 

7230 

7278 

43 

IOI7 
IO18 

7321 

7406 

7449 

7492 

7534 

7577 

7620 

7662 

7703 

43 

7748 

7790 

7833 

7876 

l%l 

7961 

8004 

8046 

8i32 

43 

IOI9 

8174 

8217 

8259 

83o2 

8387 

8430 

8472 

8558 

43 

1020 

008600 

8643 

8685 

8728 

8770 

88i3 

8856 

8893 

8941 
9366 

8983 

43 

1021  ' 

9026 

9068 

9111 

9153 

9196 

923s 

9281 

9323 

9408 

42 

1022 

9431 

9493 

9536 

9578 

9621 

9663 

9706 

9748 

9791. 

9833 

42 

1023 

9876 

X 

9961 

•ooo3 

•0043 

.0088 

•oi3o 

•0173 

•0213 

•  0258 

42 

-  1024 

oio3oo 

0385 

0427 

0470 

05l2 

o554 

0597 

0639 

io63 

0681 

42 

1020 

010724 

0766 

0809 

o85i 

0893 

0936 

0978 

1020 

no5 

42 

1026 

1147 

1 190 

1232 

1274 

i3i7 

1359 

1401 

1444 

i486 

i528 

42 

1027 

1570 

i6i3 

1655 

1697 

1740 

1782 

1824 

1866 

1909 

1931 

42 

1028 

1993 

2o3o 

2078 

2 1 20 

2162 

2204 

2247 

2289 

233 1 

2373 

42 

1029 

24i5 

2458 

2,500 

2542 

2584 

2626 

2669 

2711 

2753 

2795 

42 

io3o 

012837 

2879 

2922 

2964 
3385 

3oo6 

3048 

3090 

3i32 

3174 

3217 

42 

io3i 

3259 

33o. 

3343 

3427 

3469 

35ii 

3553 

3596 

3638 

42 

I032 

368o 

3722 

3764 

3  806 

3848 

3890 

3932 
4353 

IV4 

4016 

4o58 

42 

io33 

4100 

4142 

4184 

4226 

4268 

43io 

4437 

4479 
4898 

42 

io34 

4521 

4563 

46o5 

4647 

4689 

4730 

4772 

4814 

4856 

42 

io35 

014940 
5360 

4982 

5o24 

5o66 

5io8 

5i5o 

5192 

5234 

5276 

53i8 

42 

io36 

5402 

5444 

5485 

5527 

5569 

56ii 

5653 

5695 

5737 

42 

1037 

5779 

5821 

5863 

5904 

5946 

5988 

6o3o 

6072 

6114 

6i56 

42 

io38 

6197 

6239 

6281 

6323 

6365 

6407 

6448 

6490 

6532 

6574 

42 

1039 

6616 

6657 

6699 

6741 

6783 

6824 

6866 

6908 

6950 

6992 

42 

1040 

017033 

7075 

7117 

7159 

7200 

7242 

7284 

7326 

7367 

7409 

42 

io4i 

^H 

7492 

7534 

7576 

7618 

7659 

7701 

7743 

7?84 

7826 

42 

1042 

7868 
8284 

7909 

7951 

7993 

8o34 

8076 

8118 

8139 

8201 

8243 

42 

1043 

8326 

8368 

8409 

845 1 

8492 

8534 

857b 

86.7 
9033 

8659 

42 

1044 

8700 

8742 

8784 

8825 

8867 

8908 
9324 

8950 
9366 

8992 

9075 

42 

1045 

019116 

9i58 

9199 

9241 

9282 

9407 

9449 

9490 

42 

1046 

9532 

9573 

9613 

9656 

9698 

9739 

9781 

9822 

9864 

9903 

42 

1047 
1048 

9947 

9988 

•  oo3o 

•0071 

•oii3 

•  0154 

•  0193 

.0237  i-0278 

•0320 

41 

"o2o36i 

o4o3 

0444 

0486 

0527 

o568 

0610 

o65i 

0693 

0734 

41 

1049 

0775 

0817 

0858 

0900 

0941 

0982 

1024 

io65 

1107 

1 148 

41 

io5o 

021 189 

I23t 

1272 

13,3 

i355 

1396 

1437 

1479 

l520 

i56i 

41 

io5i 

i6o3 

1644 

i685 

1727 

1768 

1809 

l83l 

1892 

1933 
2346 

1974 

41 

1032 

2016 

2057 

2098 

2140 

2181 

2222  1  2263 

23o5 

2387 

41 

io53 

2428 

2470 

25ll 

2352 

2593 

2635,  2676 

2717 

2758 

2199 

41 

io54 

2841 

2882 

2923 

2964 

3oo5  1  3047  1  3o88 
3417'  3458:  3499 

3129 

3170 

3211 

41 

io55 

023232 

3294 

3335 

3376 

3541 

3582 

3623 

41 

io56 

3664 

3705 

3746 

3787 

3828:  3870  3911 
4239  4280 ;  4i2i 

3932 

3993 

4o34 

41 

io57 
io58 

4075 

4116 

4137 

4198 

4363 

4404 

4445 

41 

4486 

4527 

4568 

4609 

4630 

4691  4732 

4773 

4814 

4855 

41 

1059 

4896 

4937 

4978 

5oi9 

5o6o 

5ioi 

5i42 

5i83 

5224 

5265 

41 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

28 


LOGAEITHMS     OF    XU^IBERS. 


m 

0 

1 

2 

3 

4 

5 

6 

7   j   8   1   9 

D. 

1060  ! 

o253o6 

5347 

5388 

5429 
5838 
6247 

5470 

55ii 

5552 

5593  !  5634  i  5674 

41 

1061  1 
1062 

5715 
6125 

5756 
6i65 

5797 
6206 

5879 
6288 

5920 
6329 

5961 
6370 

6002 
6411 

6043  ,  6084 
6452  6492 

41 
41 

1063 

6533 

6514 
6982 
7390 

661 5 

6656 

6697 

6737 

6778 
7186 

6819 

6860  6901 

41 

1064 

6942 
027350 

7023 

7064 

7io5 

7146 

7227  7268 

7300 
7716 

41 

io65 

7431 

7472 

75i3 

7553 

7594 

7635  7676 
8042  8o83 

41 

1066 

7757 

ml 

7839 

7«79 

7920 

& 

S002 

8124 

41 

1067  1 

8164 

8246 

82S7 

8327 

8409 

8449  1  8490 

853 1 

41 

1068 

8571 

8612 

8653 

8693 

8734 

8775 
9181 

88i5 

8856  1  8896  8937 

41 

1069 

8978 

9018 

9059 

9100 

9140 

9221 

9262  j  93o3 

9343 

41 

1070 

029384 

9424 

9465 

9506 

9546 

9587 

9627 

9668  !  9708 

9749 

41 

1071 

9789 

9830 

9871 

T6 

Itr, 

9992 

.oo33  •0073-0114 

-oi54 

41 

1072 

o3oi9D 

0235 

0276 

0397 

0438 

04781  0319 

o883  1  0023 
1287  1328 

0559 

40 

1073  1 

0600 

0640 

0681 

0721 

0762 

0802 

0843 

Xt 

40 

1074 

1004 

1045 

io85 

1126 

1166 

1206 

1247 

40 

1073 

o3i4o8 

1449 
1853 

189? 

i53o 

1570 

1610 

i65i 

1691 

1732 

1772 

40 

1076 

1812 

1933 

1974 
2377 

2014 

2o54 

2093 

2i35 

2173 

40 

1077 

2216 

2256 

2296 

2337 

2417 

2458 

2498 

2538 

2578 
2981 

40 

1078 

2619 

2609 

2699 

2740 

2780 

2820 

2860 

IZ 

2941 

40 

1079 

302I 

3062 

3l02 

3i42 

3182 

3223 

3263 

3343  3384 

40 

1080 

033424 

3464 

35o4 

3544 

3585 

3625 

3665 

3705 

1^43 

3786 

40 

1081 

3826 

3866 

3906 
43o8 

3946 

3986 
4388 

4027 

4067 
4468 

4107 

4147 

4187 

40 

1082 

4227 
4628 

4267 

4348 

4428 

45o8  4548 

4588 

40 

1083 

4669 

4709 

4749 

4789 

4829 

4869 

nziitPo 

it 

40 

1084 

5029 

5069 

5109 

5i49 

5190 

523o 

5270 

40 

io85 

o3543o 

5470 

55io 

5550 

5590 

5630 

5670 

5710  i  5750 

5790 

40 

1086 

5830 

5870 

6?09 

6?49 

5990 

6o3o 

6070 

6110 

6i5o 

6190 
6580 

7387 

40 

1087 

6230 

6269 

6389 

6429 
6828 

6469 
6868 

65o9 

6008 

.7307 

6549 

40 

1088 

6629 
7028 

6669 

6709 
7108 

6749 
7140 

6789 

6048 
7347 

40 

1089 

7068 

7187 

7227 

7267 

40 

1090 

037426 

7466 

7506 

7546 

7586 

7626 

7665 

7705 
8io3 

8143 

7785 
8i83 

40 

1091 

7825 
1   8223 

7865 

IToi 

m 

7984 

8024 

8064 

40 

1092 

8262 

8382 

8421 

8461  i  85oi 

8541 

858o 

40 

1093 

8620 

8660 

8700 

8739  8779 
9i36  9176 

8819 

8859 

8898  :  8o58 
9295  '  9335 

8978 
9374 

40 

1094 

9017 

9057 

9097 

9216 

9253 

40 

1093 

039414 

9454 

9493 

9533  9573 

9612 

9652 

9692  1  9731 

9771 

40 

1096 

9811 

985o 

9800 

mi    Itt 

•  0009 

.0048 

•  0088  ,-0127  1-0167 

40 

1097 

040207 

0246 

0286 

040D 

0444 

0484  1  o523 

o563 

40 

1098 

0602 

0642 

0681 

0721  0761 

0800 

0840 

0879  0019 
1274  i3i4 

0958 
i353 

40 

1099 

0998 

1037 

1077 

1116  ii56 

1195 

1235 

39 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

TABLE    II. 


CONTAINING 


NATURAL  SI?^^ES  AND  COSHES, 


LOGAEITIIMIC   SI:N^ES,  COSINES,  TAiNGENTS,  AND 
COTANGENTS, 


EVERY  DEGREE  AND  inXTTE  OF  THE  QUADRANT. 


30 


TRIGOXCniETRICAL   FUXCTIOXS.— 0°. 


Nat.  Functions. 

Logarithmic  F 

l-nctions 

+  10. 

0 

N.sine.'  N.  cos. 

L.  sine. 

D.1" 

L.  COS.  D.l" 

L.  tang. 

D.l" 

L.  cot  1 

00000  1  Unit. 

0-000000 

10.000000 1    1 

0-000000 ! 

Infinite.  |  60 

1 

ooo2g 

Unit. 

6.463726 

50I7-I7 

000000 

00 

6-463726  5oi7 

17 

13.5362741 59 

2 

00058 

Unit. 

764756  2934- 

85 

000000 

00  1 

764756 

2934 

83 

235244  58 

S 

00087 

Unit. 

940847 

2082 

3i 

000000 

00: 

940847 

2082 

3i 

059153  57 

4 

001 16 

Unit. 

7.065786 

i6i5 

JJ 

000000 

00  ■ 

7-06D786 

i6i5 

17 

12.934214  56 
837304  55 

5 

00145  1  Unit. 

162696 

i3i9 

000000 

00 

162696 

i3i9 

69 

6 

001751  Unit. 

241877 

HID 

75 

9-999999 i 

oil 

241878 

1115 

78 

758122; 54 

7 

00204 

Unit. 

308824 

966 

53 

999999 1 

01  1 

308825 

t. 

53 

601 1 75 153 

8 

00233 

Unit. 

366816 

852 

54 

999999 

01  1 

366817 

54 

633183152 

9 

00262 

Unit. 

417968 

762 

63 

999999 
999998 

01! 

417970 
463727 

762 

63 

582o3o'51 

10 

00291  i  Unit. 

463725 

"689 

88 

_0I_ 

689 

88 

536273; 50 

11 

oo32o  j  99999 

7T5051T8 

629 

81 

9.999998 

01 

7'5o5i2o 

629 

81 

12^494880  49 

12 

00349  99999 

542906 

579 

36 

999997  i 

01  , 

542909 

IV, 

33 

457091 

48 

13 

00378!  99999 

577668 

536 

41 

999997 

01 

577672 

42 

422328 

47 

14 

00407 

99999 

609853 

499 

38 

999996  i 

01 

609837 

499 

39 

390143 

46 

15 

00436 

99999 

639816 

467 
438 

14 

999996 , 

01 

639820 

467 

i5 

36oi8o 

45 

16 

00465 

99999 

667845 

81 

999995  1 

01 

667849 

438 

82 

332i5i 

44 

17 

00495 

99999 

694173 

4i3 

72 

999995 1 

01 

694179 

4i3 

73 

3o5S2i 

43 

18 

oo524 

99999 

718997 

391 

35 

999994 1 

01 

719003 

391 

36 

280997 

42 

19 

00553 

99998 

742477 

371 

\l 

999993  : 

01 ; 

742484 

371 

28 

237316 

41- 

20 

oo582 

99998 

764754 

353 

999993 

s±, 

764761 

35i 

36 

235239  ^"^ 

21 

006 1 1 

99998 

7-785943 

336 

72 

9.999992; 

01 

7-785951 

336 

"73 

12^214049  3y 

22 

00640 

99998 

806146 

321 

75 

999991  1 

01 1 

8061 55 

321 

76 

193845  33 

23 

00669 

99998 

825451 

3o8 

o5 

999990 1 

01  ' 

825460 

3o8 

06 

174540  37 

24 

00698 

99998 

843934 

'M 

47 

999989 ; 
999988 1 

02 

843944 

295 

49 

l56o56  36 

25 

00727 

99997 

861662 

88 

02 

861674 

283 

90 

138326  85 

26 

00736 

99997 

878695 

273 

17 

999988 1 

02  i 

878708 

273 

18 

121292  34 

27 

00785 

99997 

895085 

263 

23 

999987 1 

02 

895099 

263 

25 

104901 

33 

28 

00814 

99997 

910879 

253 

I^ 

999986 1 

02^ 

910894 

254 

01 

089106 

32 

29 

00844 

99996 

926119 

245 

999985 1 

•02 

926134 

245 

40 

073866 

31 

SO  i  00873 

99996 

940842 
7.955082 

237 

33 

999983 ; 

•02 

940858 

237 

35 

059142 

30 

31 

00902 
00931 

99996 

229 

80 

9.9999821 

.02 

■^  "9^500° 

229 

81 

12-044900  2y  1 

82 

99996 

96S870 

222 

73 

999981 

•02  \ 

?s 

222 

75 

o3iiii 

23 

£3 

00960 

99995 

982233 

216 

08 

999980 

•02  1 

216 

10 

017747 

27 

£4 

00989 
01018 

99993 

995108 
80077S7 

209 

81 

999979 

•02  ' 

995219 

209 

-83 

004781 

26 

35 

9999D 

203 

t 

999977 

•02 

8-007809 

203 

& 

11-992191 

25 

36 

01047 

99995 

020021 

198 

999976 1 

•02 

020045 

198 

979933 

24 

£7 

01076 

99994 

031919 

193 

188 

02 

999975  i 

•02 

031945 
043527 

•  o5 

96S055  23 

38 

oiio5  99994 

043501 

01 

999973  i 

•02 

•  o3 

956473  22 

39 

on 34  99994 

054781 

i83 

25 

999972  1 

•02 

054809 

iS3 

•27 

945191  21 

40 

01 164  99q93 

065776 

178 

Jl 

999971  1 

•02 

o658o6 

178 

•74 

934194  20 

"IT 

01193  99993 

8.076500 

174 

•41 

-»^^ 

•02 

8-076531 

174 

•44 

11 -923460] 19 

42 

01222  99993 

086965 

170 

•  3i 

•02 

086997 

170 

•34 

9i3oo3 

18 

43 

oi25i  99992 

097183 

166 

.39 

999966 

.02 

097217 

166 

•42 

002783 
8g3o37 

17 

44 

01280  99992 

107167 

162 

.65 

999964 : 

.03 

107202 

162 

.68 

16 

45 

oi3o9  99991 

116926 

159 

•  08 

999963  I 

•03 

116963 

159 

10 

15 

46 

oi338  99991 

126471 

i55 

.66 

999961  ! 

•  o3 

i265io 

l53 

.68 

873490 

14 

47 

01367  99991 

i358io 

l52 

•38 

99993Q  ; 

9999D8  j 

•  o3 

i3585i 

l52 

•41 

864149 

13 

43 

01396  99990 

144953 

149 

•24 

•o3 

144996 

149 

•27 

855004  12  1 

49  !1  014251  QQQQO 

153907 

146 

•  22 

999956 

•  o3 

ID3932 

146 

'11 

846048 ;  1 1 1 

50 

01454 i  999^9 

162681 

143 

.33 

999954 

•  o3 

162727 

143 

•  36 

85/273 

10 

51 

oi483|  99989 
oi5i3l  99989 

8.171280 

140 

•54 

9-999952 

•03 

8-171328 

140 

^ 

11-828672 

9 

52 

179713 

i37 

86 

999900 

•  03 

\llti 

i37 

.90 

820237 

8 

53 

01 542  i  99988 

187985 

i35 

29 

999948 

•  o3 

i35 

.32 

811964 

7 

54 

01571  j  99988 

196102 

l32 

80 

999946 

•  o3 

196156 

l32 

.84 

8o3844 

6 

55 

01600 1  99987 

204070 

i3o 

41 

999944 

•  o3 

204126 

i3o 

•44 

795874 

5 

56 

01629!  99987 

211895 

128 

10 

999942 

•  04 

211953 

128 

•14 

788047 

4 

57 

01653 

99986 

219581 

125 

87 

999940 

•  04 

219641 

125 

.90 

780359 

3 

58 

01687 

99vS6 

227134 

123 

72 

999938 ; 

•  04 

227195 

123 

•7^ 

772805 

2 

59 

01716 

999R5 

234557 

121 

64 

999936 

•  04 

234621 

121 

.68 

765379 

1 

60 

01745 

999S5 

241855 

119.63 

999934! -04 

241921 

119-67 

758079 

0 

N.  COS.  :N.  sine. 

L.  COS. 

D.l" 

L.  sine,  j 

L.cot. 

D.l" 

L.  tang. 

' 

89^                         1 

TRIGOXOMETRTCAL   FUXCTIOXS. — 1' 


31 


Nat.  Functions, 

Logarithmic  Functions 

+  10. 

1 

/ 

N.  sine.  N.  cos. 

L.  sine. 

D.l" 

L.  cos.  JD.l", 

L.  tang. 
8-241921 

119-67 

L.cot.  j   1 

01745  99985 

8-241855 

119  63 

9.999934 

•04! 

11-758079 

60 

1 

01774 '999'^4 

249033 

117 

68 

999932 

•04 1 

249102 

117 

72 

700898 

59 

21 

oi8o3  99984 

256094 

ii5 

80 

999929 

•041 

256i65 

ii5 

84 

743835 

53 

3' 

oi832  99983 

263o42 

n3 

98 

999927 

-04 

263ii5 

114 

02 

736885 

57 

4 

01862  99983 

269881 

112 

21 

999925 

-04 

269956 

112 

25 

730044 

66 

5i 

01 89 I  99982 

276614 

110 

5o 

999922 

.04 

276691 

no 

54 

723309 

65 

6 

01920  99982 

283243 

108 

83 

999920 

-04 

283323 

108 

ll 

716677 

54 

7 

01949  99981 
01978,99980 

289773 

107 

21 

999918 

.04 

289856 

107 

710144 

£3 

8 

296207 

io5 

65 

9999  J  5 

-04 

296292 

lOD 

70 

703708 

62 

9 

02007  999S0 

302546 

104 

i3 

999913 

-04 

302634 

104 

18 

697366 

61 

10 
11 

o2o36  99979 

308794 

102 

66 

999910 

-04 

308884 

102 

70 

691116 

50 

02065  99979 
02094  9997a 

8-314954 

lOI 

22 

9.999907 

•  04 

8-3i5o46 

101 

26 

11-684954  4^  j 

12 

321027 

99 

82 

999900 

-04 

321122 

99 

87 

678878 

48 

13 

02123  99977 

327016 

98 

47 

999902 

-04 

327II4 

98 

5i 

672S86 

47 

14 

02l52 

99977 

332924 

97 

14 

999H99 

-o5 

333025 

97 

19 

666975 

46 

15 

O2181 

99976 

338753 

95 

86 

999897 

-o5 

338856 

95 

90 

661144 

45 

16 

022II 

99976 

344504 

94 

60 

999894 

-o5 

344610 

94 

65 

655390 

44 

17 

02240  99975 

35oi8i 

93 

38 

999891 

-o5 

350289 

93 

43 

6497 'I 

43 

18 

02260,99974 
02298199974 

355783 

92 

;? 

999888 

•  o5 

355895 

92 

24 

644105 

42 

19 

36i3i5 

% 

999885 

-o5 

36i43o 

91 

08 

638570 

41 

20 
21 

02327  1  99973 

366777 

90 

999882 

-o5 
-o5 

366895 
8-372292 

89 
88 

95 

"85 

633io5 

40 

02356 

99972 

8-372171 

88 

80 

9.999879 

11.627708 

"39" 

22 

02385 

99972 

377499 

87 

72 

999876 

-o5 

377622 

87 

•77 

622378 

38 

23 

02414 

99971 

382762 

86 

67 

999873 

-o5 

382889 

86 

-72 

617111 

37 

24 

02443 

99970 

387962 

85 

64 

999870 

-o5 

388092 
393234 

85 

-70 

61 1908 

36 

25 

02472 

99969 

393101 

84 

64 

909867 

-o5 

84 

-70 

606766 

35 

26 

025oi 

99969 

398179 

83 

66 

999864 

-o5 

3983 1 5 

83 

•71 

6oi685 

34 

27 

o253o 

99968 

4o3i99 

82 

71 

999861 

.o5 

403338 

82 

t 

596662 

33 

28 

0256o 

99967 

408161 

81 

77 

999858 

-o5 

4o83o4 

81 

591696 

32 

29 

02589 
02618 

02647 

99966 

4i3o68 

80 

86 

999854 

-o5 

4i32i3 

80 

•91 

586787 

31 

30 
31 

99966 

4i79'9 

79 

it 

999851 

-06 

418068 

80 

-02 

581932 

30 

99965 

8-422717 

79 

09 

9.999848 

-06 

8-422869 

79 

~iT 

ii-577i3i 

29 

32 

02676 

99964 

427462 

78 

23 

999844 

-06 

427618 

78 

•3o 

572382 

28 

33 

02705 

99963 

432156 

77 

40 

999841 

-06 

4323i5 

77 

•45 

567685 

27 

84 

02734 

99963 

436800 

76 

57 

999838 

-06 

436962 

76 

-63 

563o38 

26 

35 

02763 

99962 

441394 

75 

77 

999834 

-06 

44i56o 

75 

-83 

558440 

25 

36 

02792 

99961 

445941 

74 

99 

999831 

-06 

446110 

75 

.05 

553890 

24 

37 

02821 

99960 

450440 

74 

22 

999827 

.06 

45o6i3 

74 

.28 

549387 

23 

88 

o285o 

99959 

454893 

73 

46 

999823 

-06 

455070 

73 

52 

544930 

22 

39 

02879 

99939 
99958 
99957 

459301 

72 

73 

999820 

-06 

459481 

72 

•79 

54o5i9 

'21 

40 
41 

02908 
02938 

463665 

72 

00 

999816 

-06 

463849 

72 

06 

536i5i 

20 

8-467985 

71 

29 

9.999812 

-06 

8-468172 

71 

.35 

11-531828 

TIT 

42 

02967  99956 

472263 

70 

60 

999809 

-06 

472454 

70 

.66 

527546 

18 

43 

02996  j  99955 

476498 

69 

91 

999805 

.06 

476693 

69 

98 

523307 

IT 

44 

o3o25 ' 99954 

480693 

^ 

24 

999801 

-06 

480892 

69 

.31 

519108 

16 

45 

o3o54 1 99953 

484848 

59 

999797 

-07 

485o5o 

68 

-65 

5i495o 

15 

46 

o3o83  99952 

488963 

67 

94 

999793 

•07 

489170 

68 

-01 

5io83o 

14 

47 

o3ii2 199952 

493040 

^ 

3i 

999790 

-07 

493250 

67 

-38 

506750 

IG 

48 

o3i4i 1 99951 

497078 

66 

^ 

999786 

.07 

497293 

66 

-76 

502707 

12 

49 

o3 1 70- '99950 

5oio8o 

66 

999782 

-07 

501298 

66 

-i5 

498702 

11 

"50 
51 

03199-99949 
03228 ' 99948 

5o5o45 
8.508974 

65 
64 

48 
"89 

999778 
9-999774 

-07 

!  505267 

65 

.55 

494733 

10 
9 

8-509200 
513098 

64 

96 

11-490800 

52 

03257:99947 
1  03286 :  99946 

512867 

64 

3i 

999769 

•07 

64 

39 

4%902 

8 

53 

516726 

63 

75 

999765 

-07 

516961 

63 

82 

483o39 

7 

54 

o33i6  99945 

52o55i 

63 

•9 

999761 

•  07 

520790 

63 

26 

479210 

6 

55  j'  o3345  !  99944 

524343 

62 

64 

999757 

-07 

524586 

62 

72 

475414 

5 

56  1  03374  99943 

528102 

62 

11 

999753 

-07 

528349 

62 

18 

47i65i 

4 

57  1  o34o3 199942 

531828 

61 

58 

999748 

-07 

532080 

61 

65 

•  467920 

8 

58  03432; 99941 

535523 

61 

06 

999744 

-07 

535779 

61 

i3 

464221 

2 

59  1 1 03461 1 99940 

539186 

60 

55 

999740 

-07 

539447 

60 

62 

460553 

1 

60  03490 ; 99939 

542819 

60-04 

999735 

•  07 

543084 

60-12 

11-456916 

0 

|.N.  COS.  N.  sine. 

L.  COS. 

D.l" 

L.sine.  1 

L.cot. 

D.l" 

L.  tang.  1  ' 

88° 

32 


TRIGOXOMETRICAL   FUXCTIOXS. — 2" 


Nat.  Fr^-cTioNS. 

LOGAKITHMIC  FUNCTIONS  +  10. 

'  [^N.sine.  N.  cos. 

L.  sine.   D.  1" 

L.  COS.  D.l" 

L.  tang.  1  D.  1" 

L.  cot. 

0 

1 
o 

l\ 

6 

'si 

9 

10 

03490  99939 
o35i9  99938 
0354S  99937 
03577  99936 
o36o6  99935 
o3635  99934 
03664  99933 
o36q3  99932 
03723  99931 
03752  99930 
03781  99929 

8-542819  60 
546422  1  59 

557054  i  58 
56o54ol  57 

570836 '  56 
5742141  55 
577566  I  55 

04 
55 
06 
58 
11 
65 

•9 

74 
3o 
87 
44 

9.999735 
999731 
999726 
999722  : 
999717 
999713 
999708 
999704 
999699 

999&89 

•07 

•07; 

:SI 

.081 
•081 
.081 
.081 
-08! 
.08 
.08 

8-5430841  60 

553817  1  58 
557336 1  58 
560828 j  57 
564291  57 
567727  56 
571137  56 
574520  55 
577877  1  55 

12 
62 
14 

66 

it 

38 

95 
02 

11-456916 
453309 
449732 
446183 
442664 
439172 
430709 
432273 
428863 
425480 
422123 

60 
69 
58 
57 
66 
55 
54 
53 
52 

% 

11 
12 
13 
14 
15 
16 
17 
18 
19 
20 

o38io  99927 
o3S39  99926 
o3S68  99925 
03897  99924 
03926  99923 
"03955  99922 
03984  99921 
04013  99919 
04042  99918 
04071 ,999 '7 

8080892 
584193 
587469 
590721 
593948 
597152 
6oo332 
603489 
606623 
609734 

55 
54 
54 
53 
53 
53 

52 
52 

5i 
5i 

02 
60 
J9 

11 

00 
61 

23 

86 
49 

9-999683 
999680 
999673 
999670 
999665 
999660 
999655 
99965o 
999645 
999640 

.08 

•08: 

.081 

.08  i 

.o8i 

-08  i 

.08: 

.081 

.09 

-09 

8-581208 
5845i4 
587795 
591051 
594283 
597492 
600677 
6o3839 
60697a 
610094 

55 
54 
54 
53 
53 
53 

52 
52 

10 

68 

I] 

47 
08 

It 

11-418792 
415486 

4l2205 

408949 

399323 

396161 

393022 

389906 

tl 

47 
46 
45 
44 
43 
42 
41 
40 

21  !;  04100  99916 

22;  04129  99915 

23  04159  99913 

24  041SS  99912 
-5  '04217  99911 
26,;  04246  99910 

27  jl  04275  99909 

28  1104304  99907 
29:04333  99906 
30;  04362  99905 

8-612823  5i 
615891  1  5o 
618937!  5o 
621962  5o 
624965  :  49 
627948  i  49 
630911  J  49 
633854 !  48 
636776 !  48 
639680 ';  48 

12 

76 
41 
06 
72 
38 
04 

11 

-06 

9-999635 
999629 
999624 
999619 
999614 
999608 
999603 
999597 
999592 
999586 

-09 
-09 

•  09 
-09 
-09 
-09 
-09 
-09 
-09 
-09 
-09 
-09 
-09 
-09 
-10 
- 10 
-10 

•  10 
-10 
-10 

8.613189'  5i 

616262 ;  5o 

619313  1  5o 
622343!  5o 
625352  !  49 
628340 ;  49 
63i3o8'  49 
634256  48 
637184  48 
640093  48 

21 
85 
5o 
i5 
81 

47 
i3 
80 
48 
16 

11-386811 
383738 
380687 

374648 
371660 
368692 
365744 
362816 
359907 

39 
38 
37 
86 
85 
84 
83 
82 
31 
80 

31 
32 
33 
34 
35 
36 
87 
88 
39 
40 

, 04391  , 99904 
1 04420  99902 

:  04449  99901 
04478  99900 
04507  99898 

,04536  99897 
; 04565 ; 99896 

: 04594  99S94 

; 04623  99893 
04653  99S92 

8-642563 
645428 
648274 
65iio2 
65391 1 
656702 
659475 
662230 
664968 
667689 

47 
47 

46 
46 
46 
45 
45 
45 
45 

12 

82 

•52 

-22 

21 

35 
06 

9-999581 
999575 
999570 
999564 
999558 
999553 
999547 
999541 
999535 
999529 

8-642982 
645853 
648704 
65i537 
654352 
657149 
659928 
662689 
665433 
668160 

V 
47 

47 

46 
46 
46 
46 
45 
45 
45 

-84 
53 
22 

t\ 

3i 
02 

73 
44 
26 

11.357018  29 
354147  28 
351296  27 
348463  ;  26 
345648  1  25 
342851  ;  24 
340072  ,  23 
337311!  22 
334567  ■  21 
331840  20 

41 

42 
43 

44 
45 
46 
47 
4S 
49 
50 

04682  ^  99890 
,04711  i  99889 
i  04740 !  99888 
04769  99886 
0479S  I  99S85 
04827  1  99883 
; 04856  99882 
04885  99881 

; 04914 '99879 
1 04943  99878 

675751 

678405 
681043 

683665 
686272 
688863 
691438 
693998 

44 
44 
44 
43 
43 
43 
43 
42 
42 
42 

:?? 

•24 

97 
-70 
•44 
•18 
-92 
-67 
•42 

9-999524 
999518 
999512 
999006 
999500 
999493 
999487 
999481 
999475 
999469 

•  10 
-10 
•10 
-10 
•10 
-10 
•10 
- 10 

•  10 
-10 ' 

8-670870  44 
D73563  44 
676239  44 
678900  44 
681544  43 
684172  43 
686784 '  43 
689381  i  43 
691963  '  42 
694029  42 

88 
61 
34 

'bI 

54 
28 
o3 

11 

n.329i3o,19 
326437  1 18 
323761 1 17 
32II00  16 
3i8456  15 
3i5828  14 
3i32i6  13 
310619  12 
3o8o37 ; 11 
3o547i j 10 

51 

52 
53 
54 
55 
56 
57 
58 
59 
60 

1 04972  99876 
o5ooi  99875 
o5o3o  99873 
o5o59  99872 

;o5o88  99870 
o5ii7  99869 
o5i46  «99S67 

,o5i75  99866 
o52o5  99S64 

io5234  99863 

8-696543 
699073 
701589 
704090 
706577 
709049 
7ii5o7 
713902 
716383 
718800 

42 
41 
41 
41 
41 
40 
40 
40 
40 
40 

17 

9^ 
68 

44 

21 

97 

74 

-5i 
29 

-06 

9-999463 
909406 
999450 
999443 
999437 
999431 
999424 
999418 
9994" 
999404 

-11 

-11 

•II  . 

-11 

•II 

-II 

-II 

-II  , 

8-697081  42 
699617,  42 
702139  41 
704646  41 
707140 1  41 
709618;  41 
712083  :  40 
714534  40 
716972  40 
719396  40 

28 
o3 

It 

32 

08 

85 
62 
40 

■is 

297861 
295354 
292860 
290382 
287917 

285465 
283028 
280604 

9 
8 
T 

6 
5 
4 
8 

2 

1 
0 

■  N.  COS.  N.  sine. 

L.  COS. 

D.  1"  1  L.  sine. 

1 

L.cot.  i  D.l" 

L.  tang. 

/ 

§7°                         1 

TRIGO:S"OMETRIC AL   FUKCTIOXS.  — 3°. 


33 


Nat.  Functions. 

LooAKiTHMic  Functions  +  10.             1 

0 
1 

2 
S 

41 

I 

7 
8 

,1 

11! 

12  1 

\l 

15 
16 
17 
18 
19 
20 

N.sine.'N.  COS. 

L.  6ine. 

D,  1"   L.  COS.  ] 

D.l" 

L.  tang. 

D.  1"  1  L.  cot. 

05234 
05263 
09292 
05321 
o535o 
05379 
05408 
05437 
05466 
05495 
o5524 

99863 
99861 
99860 
99858 
99857 
99855 
99854 
99892 
99831 
99849 
99847 

8.718800 
721204 
723595 
725972 
728337 
730688 
733027 
735354 
737667 

739969 
742259 

40-06 
39-84 
39-62 
39-41 
39-19 
38-98 
38-77 
38-57 
38-36 
38-i6 
37-96 

9.999404 
999398 
999391 
999384 
999378 
999371 
999364 
999337 
999330 
999343 
999336 

•II 
•II 
•II 
•II 
•II 
-II 
-12 
•12 
•12 
•12 
•12 

8^719396 
721806 
724204 
726588 
728959 
73i3i7 
733663 

740626 

742922 

8^745207 

747479 
749740 
751989 
754227 
736453 
758668 
760872 
763o65 
765246 

40-17 
39-95 

im 

39-30 

3? -89 
38-68 
38-48 
38-27 
38-07 

11-280604 
278194 
273796 
273412 
271041 
2686S3 
266337 
264004 
2616S3 
259374 
237078 

60 
59 
6S 
67 
60 
65 
64 
53 
52 
51 
50 
49 
43 
47 
46 
45 
44 
43 
42 
41 
40 
89 
88 
87 
86 
85 
34 
83 
82 
31 
80 

05553 
o5582 
o56ii 
o564o 
05669 
05698 

Zd 

05785 

o58i4 

99846 
99844 
99842 
99841 
99839 
9983a 
99836 
99834 
99833 
99831 

8-744536 
746802 
749055 
751297 
753528 
755747 
757955 
760131 
762337 
764511 

^^56 
37-37 
37-17 
36-98 
36-79 
36-6i 
36-42 
36-24 
36 -06 

9.999329 
999322 
9993 1 5 
999308 
999301 
999204 
999286 
999279 
999272 
999265 

•12 
•12 
•12 
•12 
•12 
-12 
-12 
-12 
-12 
•12 

37-87 
37-68 

37-49 
37.29 
37-10 
36-92 
36-73 
36-55 
36-36 
36-i8 

ii^254793 

252521 

230260 
24801 1 
245773 
243547 
24i332 
239128 
236935 
234754 

21 
22 
23 
24 
25 
26 
27 
28 
29 
80 
31 
32 
33 
84 
85 
36 
37 
38 
39 
40 

05844 
05873 
05902 
05931 
05960 
05980 
06018 
06047 
06076 
o6io5 

99829 
99827 
99826 
99824 
99822 
99821 
99819 
99817 
99815 
99813 

8-766675 
768828 
770970 
773ioi 
775223 
777333 
779434 
781524 
7836o5 
785675 

35-88 
35-70 
35.53 
35-35 
35.18 
35-01 
34-84 
34-67 
34-5i 
34-3i 

9.999237 
999230 
999242 
999233 
999227 
999220 
999212 
999205 
999197 
999189 

•12 
•l3 
-l3 

-13 
-13 

•  13 

•  i3 

•  i3 

•  13 

•  i3 

8^767417 
769578 
771727 
773866 
773995 
778114 
780222 
782320 
784408 
786486 

36-00 
35-83 
35-65 
35-48 
35-31 
35-14 
34-07 
34-80 
34-64 
34-47 

11 •232583 

230422 

228273 
226134 
224005 

221886 
219718 

217680 
215392 
2i35i4 

061 34 
06 1 63 
06192 
06221 
o625o 
06279 
o63o8 
1  06337 
06366 
06395 

99812 
99810 
99808 
99806 
99804 
99803 
99801 
99799 
99797 
99795 

8.787736 
789787 
791828 
793859 
795881 
797894 
799897 
801892 
803876 
8o5S52 

34-i8 
34-02 
33-86 
33-70 
33-54 
33-39 
33-23 
33.08 
32.93 
32-78 

9.999181 
999174 
999166 
999158 
9991 5o 
999142 
999134 
999126 
999118 
999  no 

•  13 

•  i3 
•i3 

•  i3 

•  i3 
-13 

■M 

•  13 

•  i3 

8-788554 
790613 
792662 
794701 
796731 
79S752 
800763 
802765 
804758 
806742 

34-3i 
34-i5 

33-68 
33-52 
33-37 

33-22 

33-07 
32-92 

11-211446 

2093S7 
207333 

2o52Q9 

208209 
201248 

190242 
193268 

29 
23 

26 
25 
24 
23 
22 
21 
20 
19 
18 
17 
16 
15 
14 
13 
12 
11 
10 

~9" 
8 
7 
6 
5 
4 
8 
2 
1 
0 

41 
42 
43 
44 
45 
46 
47 
48 
49 
'50 
51 
52 
53 
54 
55 
56 
57 
58 
59 
60 

1 06424 
06453 
06482 
o65ii 
06540 
06569 
06598 
06627 
06656- 
06685 

99793 
99792 
99790 
99788 
99786 
99784 
99782 
99780 
99778 
99776 

8-807819 
809777 
811726 
813667 
815599 
817522 
819436 
821343 
823240 
825i3o 

32-63 
32.49 
32.34 
32.19 
32.05 
3i  .91 
31.77 
31.63 

31.49 
31.35 

9.999102 

999086 
999077 
999069 
999061 
999053 
999044 
999036 
999027 

•14 
•14 
•14 
•14 
•14 
•14 
•14 
•14 

8-808717 
810683 
812641 
814589 
816529 
818461 
820384 
822298 
824205 
826103 

32-78 
32-62 
32-48 
32-33 
32-19 
32-o5 
31-91 
3i-77 
31-63 
3i-5o 

Il'l9i283 

i8q3i7 
1S7359 
,i3o4ii 
1 3347 1 
181539 
179616 
177702 
173795 
173897 

.06714 
06743 
06773 
06802 
o683i 
06860 
06889 
06918 
06947 
06976 

99774 
99772 
99770 
99768 
99766 
99764 
99762 
99760 
99758 
99756 

8-827011 
828884 
830749 
832607 
834456 
836297 
838 i3o 
839956 
841774 
843585 

3l.22 

3i.o8 
30.95 
30.82 
30.69 
30.56 
30.43 
3o.3o 
30.17 
3o.oo 

9.999019 
999010 

998993 
998984 
998976 

&^l 

998930 
998941 

•14 
•14 
•14 
•14 
•14 
•14 

•  i5 

•  15 

•  i5 

•  i5 

8-827992 

i  829874 

831748 

!  8336i3 

1  835471 

837321 

839163 

840998 

842825 

844644 

3i-36 
31-23 
3i-io 
30-96 
3o-83 
30-70 
30-57 
3o-45 
30-32 
30-19 

ii^i720oS 
170126 
108202 
166387 
164529 
162679 
160S37 
159002 
107175 
155356 

,N.  COS.  |N.  sine. 

L.  COS. 

D.  1"  '  L.  sine.  |    |;  L.  cot.   D.  1" 

L.  tang. 

86° 

u 


TRIGONOMETRICAL   FUXCTIOXS. — 4' 


Nat.  Functions. 

Logarithmic  Functions 

+  10. 

1 

0 

N.8ine.|N.cos. 

L.  sine. 

D.l" 

L.  COS. 

D.l" .  L.  tang. 

Dl." 

L.  cot 

06976  99756 

8-843585 

3o-o5 

9-998941 

.i5  8.844644 

30.19 

11-155356 

60 

1 

07005 1 99754 

845387 

29-92 

998932 

•10 

846455 

3o-07 

153545 

59 

2 

' 07034 '99752 

847183 

29-80 

99S923 

•  15 

848260 

29-95 

i5i74o 

5S 

8 

] 07063  9975o 

848971 

29-67 

998914 

.i5 

85oo57 

29-82 

149943 

57 

4 

:  07092 !  99748 

85o75i 

29-55 

998905 

•  i5 

851846 

29-70 

148154 

56 

5 

071 21 [99746 

852525 

29-43 

998896 

.i5 

853628 

29-58 

146372 

55 

6 

07 1 5o' 99744 

854291 

29-31 

« 

.i5 

855403 

29-46 

144597 

54 

7 

07179  99742 
07208,99740 

856049 

29-19 

•  15 

857171 

29-35 

142829 

53 

8 

857801 

29-07 

998869 

•  15 

858932 

29-23 

141068 

52 

9 

! 07237  99738 

859546 

28.96 

998860 

•  15 

860686 

29. u 

j393i4 

51 

10 
11 

[07266! 99736 

861283 

2S-84 

998851 

•  15 

862433 

29-00 

137567 

50 

107295 199734 

8-863014 

28-73 

9.998841 

•  15 

8-864173 

28-88 

11 -135827 

49 

12 

107324 

99731 

864738 

28-61 

998832 

.i5 

865906 

28.77 

134094 

43 

IS 

07353 

99129 

866455 

28 -50 

998823 

•  16 

867632 

28-66 

132368 

47 

14 

07382 

99727 

868160 

28-39 

998813 

.16 

869351 

28-54 

I 30649 

46 

15 

0741 1 

99725 

869868 

28-28 

998804 

.16 

871064 

28-43 

128936 

45 

16 

07440 

99723 

871565 

28.17 

998795 

.16 

872770 

28-32 

127230 
125531 

44 

17 

07469 
07498 

99721 

873255 

28-06 

998785 

.16 

874469 

28-21 

43 

IS 

99719 

874933 

27.95 

998776 

.16 

876162 

28-11 

123838 

42 

19 

07027  99716 

876615 

27-86 

998766 
998757 

.16 

■877849 

28-00 

122l5l 

41 

'20 

07506199714 

8782S5 

27-73 

.16 

879029 

27-89 

120471 

40 

21 

07085,99712 

8.879949 

27-63 

9.998747 

^76 

8-881202 

27.79 

11-118798 

39 

22 

07614  99710 

881607 

27.52 

998738 

.16 

882869 

27.68 

II7131 

33 

2S 

07643 

99708 

883258 

27.42 

998728 

.16 

884530 

27.58 

1 1 5470 

37 

24 

07672 

99705 

884903 

27.31 

998718 

.16 

8861 85 

27-47 

ii38i5 

36 

25 

07701 

99703 

886542 

27-21 

998708 

.16 

887833 

27.37 

112167 

85 

26 

07730 j 99701 

888174 

27-11 

998689 

.16 

889476 

27.27 

110524 

34 

27 

07759  99699 

889801 

27-00 

.16 

891112 

27.17 

108888 

33 

28 

07788 

99696 

891421 

26-90 

998679 

.16 

892742 

27-07 

107258 

32 

29 

07817 

99694 

893035 

26-80 

998669 

•17 

894366 

26.97 

105634 

31 

30 

07846 

99692 

894643 

26-70 

998609 

•17 

895984 

26-87 

104016 

30 

'si 

07875 

99689 

8-896246 

26-60 

9.998649 

•17 

8-897596 

26.77 

11-102404 

"29" 

32 

07904  99687 

897842 

26-51 

99S639 

•17 

899203 

26.67 
26.58 

100797 

28 

33 

07933  99685 

899432 

26-41 

998629 

•n 

900803 

099197 

27 

34 

07962 ; 99683 

901017 

26-31 

998619 

•17 

902398 

26.48 

097602 

26 

35 

07991  j  99680 
08020 1 99678 

902596 

26-22 

998609 

•17 

903987 

26.38 

096013 

25 

36 

904169 

26-12 

998599 

•n 

905570 

26.29 

094430 

24 

37 

08049  99676 

905736 

26-03 

998589 
998078 

•17 

907147 

26.20 

092853 

23 

38 

08078  99673 

907297 
908803 

25-93 

25-84 

•17 

908719 

26-10 

091281 
089715 

22 

39 

,08107,99671 

998568 

•17 

910285 

26-01 

21 

40 
41 

,o8i36' 99668 

910404 

25.75 

998558 

•17 

911846 

25-92 

0881 54 

20 

:o8i65j 99666 

8-911949 

20.66 

9.998548 

•17 

8-913401 

25-83 

11.086599 

19 

42 

'08194  99664 

913488 

25-56 

998037 

•17 

914951 

25.74 

o85o49 

18 

43 

; 08223  99661 

9l5022 

25-47 

998527 

•n 

916495 

25.65 

o835o5 

17 

44 

08252 199659 

9i655o 

25-38 

998516 

.18 

918034 

25.56 

081966 

16 

45 

,  08281  , 99657 

918073 

25-29 

998506 

.18 

919568 

25.47 

080432 

15 

46 

'o83io' 99654 

919591 

25-20 

998495 
998485 

.18 

921096 

25-38 

IfX 

14 

47 

08339 : 99652 

921103 

25-12 

.18 

922619 

25-30 

13 

48 

08368 1 99649 

922610 

20-03 

998474 

.18 

924136 

25.21 

075864 

12 

49 

08397  99647- 

924112 

24-94 

24-86 

998464 

.18 

920649 

25-12 

074351 

11 

50 

08426 '  99644 

925609 

998453 

.18 

927156 

25. o3 

072844 

10 

TT 

08455  i  99642 

8-927100 

24-77 

9.998442 

.18 

8-928658 

24.95 

11-071342 

9 

52 

08484  99639 

928587 

24-69 

998431 

.18 

93oi55 

24.86 

069845 

8 

53 

o85i3  99637 

930068 

24-60 

998421 

.18 

931647 

24-78 

068353 

7 

54 

08542  99635 

931 544 

24-52 

998410 

.18 

933 1 34 

24-70 

066866 

6 

55 

: 08571  99632 

9330 1 5 

24-43 

l& 

.18 

934616 

24-61 

065384 

5 

56 

08600 , 99630 

934481 

24-35 

.18 

936093 

24-53 

063907 

4 

57 

08629  99627 

935942 

24-27 

998377 

.18 

,  937565 

24-45 

062435 

3 

58 

oS658  99625 

937398 

24.19 

998366 

.18 

;  939032 

24-37 

060968 

2 

59 

08687  99622 

938850 

24-11 

998355 

-18 

I  940494 

24 -30 

059506 
o58o48 

1 

60'  08716; 99619 

940296 

24-03 

998344 

.18 

}  941952 

24-21 

0 

'  N.  COS.  N.  sine. 

L.  COS. 

D.l" 

L.  sine. 

L.  cot. 

D.l" 

L.  tang. 

~^ 

85°                         1 

TRIGONOMETRICAL  FUXCTIOXS. 


35 


Nat.  Functions. 

Logarithmic  Functions 

+-  10. 

1 
j 

'  ! 

N.sine.'  N.  cos. 

L.  sine. 

D.  1" 

L.  COS.  jD.l"; 

L.  tang. 

D.1- 

L.  cot 

^\ 

08716  !  99619 

8-940296 

24-o3 

9.998844  -19 '8.941952 

24-21 

ii^o58o48 

60 

1 

08745  Qq6i7 

941738 

28-94 

998333 : 

•19' 

948404 

24-13 

056596  59  1 

2 

08774 

99614 

943174 

28-87 

998822 1 

-19, 

944852 

24-05 

o55i48;5S  1 

3' 

o88o3 

99612 

944606 

28-79 

9988.1 1 

•19: 

946295 

23.97 

o587o5 

57 

4; 

o883i 

99609 

946034 

28-71 

998800 

-19 

947734 

23-90 

052266 

66 

5 

08860 

99607 

947456 
948874 

23-68 

998289 

-19 

949  J 68 

28-82 

o5o832 

55 

6 

08889 

99604 

23-55 

998277 : 

-19 

930597 

28.74 

049403 

54 

7 

08918 

99602 

950287 

23-48 

998266 

•19 

952021 

28-66 

047979 

53 

8 

08947 

99399 

931696 

28-40 

998255 1 

-,9 

958441 

28-60 

046559 

52 

9 

08976 

99396 

953100 

28.82 

998248 

-19 

954856 

28-5i 

045144 

51 

10 

u 

09005 

99594 

_954499 
8-955894 

28-25 

998282 

liil 

956267 

23-44 

048788 

50 

090841  99591 

28-17 

9-998220 

•I9i 

8-957674 

28.37 

u-042326 

ly" 

12 

09063  99388 

957284 

23-10 

998209 

.,9 

959075 

23.29 
28.28 

040925 

4S 

13 

09092  99386 

958670 

28-02 

.998 197 

.191 

960473 

089527 

47 

14 

091 21  1  99583 

960052 

22-95 

998186 

.19I 

961866 

23i4 

088184 

46 

15 

09150  99380 

961429 

22-88 

993174 

.19 

968255 

23-07 

086745 

45 

16 

Z'^ 

99578 

962801 

22-80 

998168 

•19 

964689 

23-00 

o3586i 

44 

17 

99373 

964170 

22.73 

998151 

.19 

966019 

22-98 

22-86 

088981 

43 

18 

09237 

99572 

965534 

22-66 

998189 

.20 

967394 
968766 

082606 

42 

19 

09266  99370 

966893 

22-59 

998128 

.20 

22-79 

081234 

41 

20 

09295 1 99367 

968249 

22-32 

998116 

.20! 

970188 

22-71 

029S67 

40 

21  i 

09324 

99564 

8-969600 

22-44 

9.998104 

.20 

8-971496 

22-65 

11^028504 

oU 

22 

09353 

99562 

970947 

22-38 

998092 

-20 

972855 

22.57 

027145 

SS 

23 

09382 

99559 

972289 

22-31 

998080 

-20 

974209 

22.51 

025791 

37 

"24' 

o'J4i  I 

99556 

978628 

22-24 

998068 

-20 

975560 

22.44 

022440 

36 

25 

09440 

99553 

974962 

22.17 

998056 

-20 

976906 

22.37 

022094 

35 

26 

011469 

99531 

976293 

22-10 

998044 

-20 

978248 

22.30 

021752 

34 

27 

09498 

99548 

977619 

22.08 

998082 

.20 

979586 

22.23 

020414 

83 

23 

09527 

99545 

978941 

21-97 

998020 

•20 

980921 

22-17 

019079 

32 

29 

09556 

99542 

980259 

21-90 

99S008 

•20 

982251 

22-10 

017749 

01 

80 
81 

09585  i  99540 

981573 

21.88 

997996 

•20 

988577 

22-04 

016428 

SO 

09614 

99537 

3-982883 

21-77 

9.997984 

-20 

8-984899 

21-97 

ii^oiSioi 

29 

82 

09642 

99534 

984189 

21.70 

997972 

•20 

986217 

21.91 
21.84 

018788 

2S 

83 

09671 

99531 

985491 

21.68 

997939 

-20 

987532 

012468 

27 

84 

I  09700 

99528 

$Vot'3 

21.57 

997947 

-20 

988842 

21.78 

011158 

26 

85  ij  09729 

99526 

21.50 

997935 

•21 

990149 

21-71 

ooo85i 

25 

86 

1 °9758 

99523 

989374 

21.44 

997922 

-21 

991401 

21^65 

008549 

24 

87 

109787   99320 

990600 

21.88 

997010 
997897 
997883 

•21 

992730 

21^58 

007250 

23 

88 

j 09816  99317 

991943 

21.31 

.21 

994045 

21^52 

005955 

22 

89 

1098451  99514 

998222 

21.25 

.21 

995337 

21^46 

004663 

21 

40 
41 

09874  995 n 
09903  99508 

994497 
8-995768 

21-19 

997872 

-21 

996624 

21^40 

21^34 

008876 

20 

21.12 

9.997860 1 

•21 

,8 -997908 

11 -002092 

TT 

42 

09932  99006 

997086 

21.06 

997847 

•21 

1  999 '88 

21.27 

000812 

IS 

43 

09961 

995o3 

998299 

21-00 

997835 

•21 

9-000465 

21.21 

10-999535 

17 

44 

09990 

99300 

999560 

20-94 

997822 

•21 

001788 

2I.l5 

998262 

16 

45 

I00I9 

99497 

9-000816 

20-87 

997809 

-21 

008007 

21^03 

996998 

15 

46 

10048 !  99494 

002069 

20-82 

997797 

-21 

004272 

995728 

14 

47 

10077!  99491 

oo33i8 

20-76 

997784 

-21 

005534 

20.97 

994466 

13 

48 

' 10106 

99488 

004568 

20-70 

997771 

•21 

006792 

20-91 

20-85 

993208 

12 

49 

;ioi35 

99485 

oo58o5 

20-64 

997758 

-21 

008047 

991933 

11 

50 
51 

IOI64 

99482 

007044 

20-58 

997745 

•21 

009298 

20-80 

?9^7_o^ 

10 

IOI92  99479 

9-008278 

20-52 

9-997782 

•21 

'9-010546 

20-74 

10 •989454 

9 

52 

! 10221  I  99476 

009510 

20-46 

997719 

-21 

011790 
018081 

20-68 

980210 

3 

53 

10230 

99473 

010787 

20-40 

997706 

-21 

20-62 

986969 

7 

54 

10279 

99470 

011962 

20-34 

997698 

•22 

014268 

20-56 

983782 

6 

55 

io3o8 

99467 

018182 

20-29 

997680 

-22 

oi55o2 

20^5l 

984498 

5 

56 

10337 

99464 

014400 

20-28 

997667 

•22 

016782 

20^45 

988268 

4 

57 

10366 

99461 

oi56i3 

20-17 

997654 

-22 

017959 

20^40 

982041 

3 

58 

10895 

99458 

016824 

20-12 

997641 

-22 

019188 

20^33 

980817 

2 

59 

10424 

99455 

oi8o3i 

20-06 

997628 

•22 

020408 

20-28 

^??3S 

1 

60 

10453 

99452 

019235 

20-00 

997614 

•22 

021620 

20-28 

0 



|n.C09,|N.  sine. 

L.  COS. 

D.  1" 

L.  sine. 

L.  cot. 

D.  1" 

L.  tang.  1  ' 

84° 

36 


TRIGOKOMETRICAL   FUNCTIONS.— 6^ 


Nat.  Functions. 

Logarithmic  Functions  +  10.             1 

1 
2 
3; 

t! 
l\ 

— i 
11 
121 
13 
14 
15 

17 
13 
19 
20 
21 
22 
23 
24 
25 
26 
27 
23 
29 
30 
81 
82 
8':> 
84 
35 
36 
87 
83 
89 
40 
41 
42 
43 
44 
45 
46 
47 
48 
49 
50 
51 
52 
53 
54 
55 
5ti 
57 
53 
59 
60 

N.  sine.  N.  cos. 

Ksine.  1  D.l" 

L.C08.  ,] 

D.l" 

L.  tang. 

D.1'' 

L.  cot 

10453  99452 
10482  99449 
io5ii  99446 
io54o  99443 
10569  99440 
10597  99437 
10626  99434 
io655  99431 
10684  99428 
10713  99424 
10742,99421 

9-019235 
020435 

02l632 

022825 
024016 

025203 

026386 
027567 
028744 
029918 
031089 

20-00 
19.95 

-9-89 
19-84 
19.78 
19.73 
19.67 
19-62 
19-57 
19-51 
19-47 

9-997614 
997601 
997588 
997374  j 
997361 
997347  1 
997334  j 
997020 
997007 
997493 
997480 

-22 

•22 

.22 

•22 

.22 

•22 

.23 

•23 

.23 

•231 

.23 

9-021620 
022834 
024044 

02525l 

026455 
027655 
028852 
o3oo46 
o3i237 
032425 
o336o9 

20 

20 
20 
20 
20 
19 
19 

^9 
19 
19 
19 

23 

17 
II 
06 
00 
95 
90 
»5 
79 

10.978380 

972345 
971148 
969954 

966391 

60 
59 
58 
57 
56 
55 
54 
53 
52 
51 
50 

10771  99418 
10800  994 1 5 
10820  99412 
io858  99409 
10887 .99406 
10916  1  99402 
10945  99399 
10973,99396 
11002  99393 
iio3i  i  99390 

9-032257 
033421 
034582 
035741 
036896 
o38o48 
039197 
040342 
041485 
042625 

19-41 

19-36 
i9-3o 
19-25 
19-20 
19.10 
19-10 
19.05 
18.99 
18-94 

9-997466 
997452 
997439 
997425 
99741 1 
997397 
997383 

997341 

•23 

-.11 

-33, 

•  23 
•23 

9.034791 
035969 
037144 
o383i6 
039485 
04065 1 
04i8i3 
042973 
O44i3o 
045284 

19 

19 
19 
19 
19 
19 
19 
J9 
19 
19 

64 

58 
53 
48 
43 
38 
33 
28 

23 

18 

10-965209 
96403  I 

962856 
961684 

9605 1 5 

%!!, 

957027 
955870 
954716 

49 

43 

47 

46 

45 

44 

43 

42 

41 

40 

"39" 

38 

37 

36 

35 

34 

83 

32 

31 

30 

1 1060  99386 
11089 '99383 
11 118! 99380 

1 1147  99377 
II 176  99374 
ii2o5'99370 
11 234 1 99367 
11263 ! 99364 
11291,99360 
1 1 320  99357 

9-043762 
044895 
046026 
047154 
048279 
049400 
o5o5i9 
o5i63d 
052749 
053859 

18.89 
18.84 
18-79 
18-75 
18.70 
i8-65 
18-60 
18-55 
18. 5o 
18-45 

9-997327  1 
9973i3 
997299 
997285 
997271 
997257 
997242 
997228 
997214 
997199 

•34 
•24 
•24 
•24 
•24 
-24' 
•24' 
•24  i 

9-046434 
047382 
048727 
049869 
001008 
052144 
053277 
004407 
o55o35 
056659 

19 
19 
19 

18 
18 
18 
18 
18 
18 
18 

i3 

08 
o3 
98 
93 
89 
84 
79 
74 
70 

10-953566 
952418 
901273 
90oi3i 
948992 
947806 
946723 
945093 
944465 
943341 

11340 '99354 
1 13-8; 99351 
11407:99347 
1 1 1436 199344 
11 1465 1 99341 
11494,99337 

1ID23  99334 

11552  99331 

iii58o! 99327 

1 1609! 99324 

Ti638!9932o 

1 1667  99317 

1 1696  99314 

11725:99310 

11754:99307 

117831 993o3 

11812  1  99300 

11840199297 

j 11869! 99293 

j 11898199290 

111927  99286 

' 11956.99283 

1  1 1985  99279 

:  12014 1  99276 

12043 ' 99272 

12071 1  99269 

! 12100' 99265 
i 1 21 29' 99262 

ji2i58  99258 
12187  99255 

9-054966 
056071 
057173 
o5827i 
059367 
060460 
o6i55i 
062639 
063724 
064806 

i8-4i 
18-36 
i8-3i 
18-27 
18.22 
18.17 
18. i3 
18.08 
18.04 
17.99 

9-997185 
997170 
997156 
997141 
997127 
997112 
997098 
997083 
997068 
997053 

•24 
•24! 

•24' 

-24 
•24' 

.25 
.25 
.25 

9-057781 
058900 
060016 
061 i3o 
062240 
063348 
064453 
965556 
066655 
067752 

18 
18 
18 
18 
18 
18 
18 
18 
18 
18 

65 
69 
55 
5i 
46 
42 
37 
33 
28 
24 

iu-942219 
941 100 

939984 
938S70 

935547 
934444 
933345 
932248 

29 
28 
27 
26 
25 
24 
23 
22 
21 
20 

9-065885 
066962 
o68o36 
069107 
070176 
071242 
072306 
073356 
074424 
075480 

9-076533 
077583 
078631 
079676 
080719 
081759 
082797 
083832 
084864 
080894 

17.94 
17-90 

17.86 
17.81 

n-77 
17-72 
17-68 

17-63 

^■M 

17-50 
17-46 
17-42 
17-38 
17-33 
17-29 
17-20 
17.21 

9-997039 
997024 
997009 
996994 
996979 
996964 
996949 
996934 
996919 
996904 

.25 

.25 
.25 
.25 
.25 
.25 

9-U68846 
069938 
071027 
072113 
073197 
074278 
075306 
076432 
0775o5 
078576 

18 
18 
18 
18 
18 

17 
17 

17 

10 
06 
02 

9'3 

80 

10.931154 
930062 
928973 
927887 
926803 
925722 
924644 
923568 
922495 
921424 

19 
IS 
17 
16 
15 
14 
13 
12 
11 
10 

9-996889 
996874 
996858 
996843 
906828 
996812 
996797 
996782 
996766 
996751 

•25 
.20 
.25 
.20 
.25 

.26 
.26 
.26 
.26 
•  26 

9.079644 
080710 

081773 

082S33 

1  083891 

;  084947 
080000 
j  087050 
,  088098 
1  089144 

17 

'7 
17 
17 

17 
17 
17 
17 
17 

76 

V 

'A 
?! 

01 

.38 

10-920356 
919290 
918227 
917167 
916109 
9i5o53 
914000 
912950 
91 1902 
910856 

9 
8 

7 
6 
5 
4 
3 
2 
1 
0 

N.  COS.  N.  sine. 

L.  COS. 

D.l" 

L.  sine. 

L.COL 

D.l" 

L.  tang. 

/ 

I 

TRIGONOMETRICAL   FUNCTIONS. — 7°. 


37 


Nat.  Functions. 

Logarithmic  Functions  +  10.             1 

1 

2 
8 
4 
5 

?! 

81 

91 

10 

11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
80 

N.sine. 

N.  COS. 

L.  sine. 

D.1" 

L.  COS.  ,] 

D.l" 

L.  tang. 

D.l"  1  L.  cot. 

12187' 
12216 
12245! 
12274 

I2302  ' 

i233i 
12360 
12389' 
12418 

12447' 
12476, 

99235 
99231 

99248 
99244 
99240 
99237 
99233 
99230 
99226 
99222 
99219 

9-085894 
086922 
087947 
088970 
089990 
091008 
092024 
093037 
094047 
093056 
'096062 

.   17 

17 
17 
16 
16 
16 
16 
16 
16 
16 

i3 

09 
04 
00 
96 
92 
88 
84 
80 
76 
73 

9.996731 
996733 
996720 
996704 
996688 
996673 
996657 
996641 
996625- 
996610 

>  .  996594 

.26 
.26 
.26 
•26 
•26 
.26 
•36 

.26 
.26 
•26 
•36 

9.089144 
090187 
091228 
092266 
093302 
094336 

^tl 

097422 
098446 
099468 

17- 
17- 
17 

17 

17 

17 
17 
17 
16 

16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 
16 

38 
34 
3o 

27 
22 

\l 

11 

07 
o3 

99 

95 

84 
80 
76 

72 

^^ 

63 
61 
.58 
•54 
.50 
.46 
•43 

it 

•32 

•29 

•23 

io^9io856 
909813 

908772 
907734 

& 

904633 
9o36o5 
902578 
901554 
900532 

10^899513 
898496 
897481 
896468 
895458 
894450 
893444 
892441 
891440 
890441 

10-889444 
888449 
887457 
886467 

885479 
884493 
683509 

8S2328 

88 1 548 
880571 

60 
59 
58 
57 
56 
55 
54 
53 
52 
51 
50 
49 
48 
47 
48 
45 
44 
43 
42 
41 
40 
39 
88 
37 
30 
35 
34 
33 
32 
31 
30 

i25o4, 
12533 

12562' 

12591 
12620 
12649 
12678 
12706 
12735 
12764 
12793 
12822 

12831 
12880 
12908 
13937 
12966 
12995 

i3o24 
i3o53 

99215 
992 1 1 
99208 
99204 
99200 

99197 
99193 
99189 
99186 
99182 

9.097065 
098066 
099065 
100062 
ioio56 
102048 
io3o37 
104025 
io5oio 
105992 

-16 
16 
16 
16 
16 

j6 
16 
16 
16 
.  16 

68 
65 
61 

ll 

it 

41 

38 
34 

9-996578 
996562 

<     996546 
996530 
996514 
996498 
996482 
996465 

996449 
996433 

•27 
•27 
•27 

•27 
•27 

•27 
•27 
•27 
•27 
•27 

9-100487 
ioi5o4 
102519 
103532 
104542 
io555o 
106556 
107559 
108360 
109559 

99178 
99175 
99171 
99167 
99163 
99160 
99136 
99152 
99  J  48 
99144 

9-106973 
107951 
108927 
109901 
110873 
111842 
112809 
113774 
114737 
113698 

16 
16 
16 
16 
16 
16 
16 
16 
16 
i5 

3o 
27 

23 

19 
16 
12 
08 
o5 
01 
97 

9-996417 
996400 
996384 
996368 
9o635i 
996335 
996318 
996302 
996285 
996269 

•27 
•27 
•27 
•27 
•27 
•27 
•27 
.28 
.28 
.28 

9.110556 
iii55i 
112543 
113533 
114521 
ii55o7 
116491 
117472 
118452 
119429 

81 
32 
33 
84 
85 
86 
37 
88 
89 
40 
41 
42 
43 
44 
45 
46 
47 
48 
49 
50 
51 
52 
53 
54 
55 
56 
57 
58 
59 
60 

i3o8i 
i3iio 
i3i39 
i3i68 
i3i97 

13226 

13254 
13283 
i33i2 
i334i 

99141 
99137 
99133 
99129 
99125 
99122 
99118 
99114 
99110 
99106 

9-116656 
117613 
118067 
119519 
1 20469 
121417 

122362 

i233o6 
124248 
125187 

i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 

94 
90 

«7 
83 
80 

]t 

69 
66 
•62 

9-996252 
996235 
996219 
996202 
996185 
996168 
996131 
996134 
996117 
996100 

.28 

•28 
.28 
•28 
•28 
.28 
.28 

.23 

.28 
.28 

9.120404 
121377 
122348 
123317 
124284 
125249 
126211 
127172 
i28i3o 
129087 

16 
16 
16 
16 
16 
16 
16 

i5 
i5 
i5 

•22 

•18 

•i5 

•  u 

•  07 

•  04 

•  01 
•97 
.94 
.91 

10-879596 

878623 
877652 
876683 
875716 
874751 
873789 
872828 
871870 
870913 

29 
28 
27 
26 
25 
24 
23 
22 
21 
20 

13370 
13399 
13427 
13436 
13485 
i35i4 
13543 
13572 
i36oo 
13629 
13658 
13687 
13716 
13744 
13773 
i38o2 
1 3831 
i386o 
13889 
13917 

99102 
99098 
99094 
99091 
99087 
990S3 

99079 
99073 

99071 
99067 

9-126125 
127060 
127993 
128925 
129854 
130781 
131706 
i3263o 
133551 
134470 

i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 

.62 

:8 

•42 

-.11 

.32 
•29 

9.996083 
996066 

996049 
996032 
996015 
993998 
995980 
993963 
993946 
993928 

•29 
•29 
.29 
•29 
•29 
•29 
•29 
•29 
•29 
•29 

9-i3oo4i 
130994 
i3i944 
132893 
133839 
134784 
135726 
136667 
137605 
138542 

i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 

1^ 
•  84 

.81 

•77 

•74 

•71 

•67 

64 
•61 

58, 

10-869939 
869006 
868o56 
867107 
866161 
863216 
804274 
863333 
862395 
861438 

19 
18 
17 
16 
15 
14 
13 
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10 
~9" 

8 

7 

6 

6 

4 

3 

2 

1 

0 

99063 
99059 
99055 
99031 
99047 
99043 
99039 
99035 
9903 1 
99027 

9.135387 
i363o3 
137216 
i38i28 
139037 
139944 
i4o85o 
141754 
142655 
143555 

i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 
i5 
14 

•25 
-22 
.19 
•16 
•12 

■2 

-o3 

-00 

.96 

9.995911 
993894 
995876 
995859 
993841 
995823 
993806 
995788 
995771 
995753 

•29 
•29 
•29 
•29 
•29 
•29 
•29 
•29 
•29 
•29 

9.139476 
140409 
141340 
142269 
143196 
144121 
145044 
145966 
146885 
147803 

i5 
i5 
i5 
i5 
i5 
15 
i5 
i5 
i5 
i5 

•55 
•5i 
•48 
•45 
42 

n 

32 

11 

io-86o524 
839391 
858660 
857731 
856804 
855879 
854956 
854034 
853115 
832197 

N.  COS. 

N.sine. 

L.  COS. 

D.l" 

L.  sine. 

1  L.  cot 

D.l" 

L.  tang. 

' 

82°                          1 

38 


TRIGONOMETRICAL   FUNCTIONS. — 8" 


Nat.  Functions. 

LoGARiTUMic  Functions 

+  10. 

0 

:N.sine. 

N.  COS. 

L.  sine. 

1  D.  1" 

L.  COS. 

D.l' 

!|  L.  tang. 

1  D." 

L.  cot 

|i39i7 

99027 

9-143555 

i  14-96 

9.995753 

-3o 

9.147803 

15-26 

10-852197 1 60 

1 

1 13946 

99023 

144453 

1  14-93 

995735 

-3o 

148718 

15-23 

851282 159 

2 

1 13975 

99019 

145349 

i  14-90 

995717 

•  3o 

149632 

l5-20 

85o368 

58 

8 

14004 

990  ID 

146243 

1  14-87 

993699 

•3o 

i5o544 

15-17 

849456 

57 

4 

14033 

9901 1 

147136 

j  14-84 

995681 

.30 

i5i454 

i5-i4 

843546 

56 

5 

14061 

99006 

148026 

!  14-81 

995664 

.30 

152363 

15-11 

847637 

55 

6 

\ 14090 

99002 

148915 

i  14-78 

995646 

.30 

153269 

i5-o8 

846731 

54 

7 

14119 

98998 

149802 

14-75 

995628 

•30 

154174 

i5-o5 

845826 

53 

8 

14143 

98994 

i5o686 

14-72 

995610 

-30 

I 55077 

l5-02 

844923 

52 

9 

14177 

98990 

i5i569 

14-69 

995591 

•3o 

155978 

14-99 

844022 

51 

10 
11 

1  i42o5  '  989% 

i5245i 

14-66 

_995^73_ 

•  3o 

156877 

14-96 

843123 

50 
49 

,  I42J4I9S9S2 

9-153330 

14-63 

9-995555 

•3o 

9.157775 

14-93 

10-842225 

12 

14263198978 

154208 

14-60 

995537 

-3o 

158671 

14-90 
14-87 

841329 

48 

13 

14292  98973 

i55o83 

14-57 

995319 

•  30 

159565 

840435 

47 

U 

!  14320  i  98969 

155957 

14-54 

995501 

-3i 

160457 

14-84 

839543 

46 

15 

14349:9^965 

1 56830 

14-51 

993482 

•  31 

i6i347 

14-81 

838653 

45 

16 

1437^:98961 

157700 

14-48 

995464 

-3i 

162236 

14-79 

837764 

44 

17 

14407 1 98957 

158569 

14-45 

995446 

-3i 

i63i23 

14-76 

836877 

43 

IS 

14436 1 98953 

159435 

14-42 

995427 

-3i 

164008 

14^73 

835992 

42 

19 

14464 ! 9S948 

i6o3oi 

14-39 

995409 

-3i 

164892 

14-70 

835 1 08 

41 

20 
21 

I44q3 '  98944 

161164 

14-36 

995390 

-3i 

165774 

14-67 

834226 

40 

14522  9S940 

9-162025 

14-33 

9-995372 

-3i 

9.166654 

14-64 

10-833346 

3D 

22 

14551 I9S936 

162885 

i4-3o 

995353 

-31 

168409 

i4-6i 

832468 

38 

23 

14580^98931 

163743 

14-27 

995334 

-3i 

14^58 

83i59i 

37 

24 

14608  :  9S927 

164600 

14-24 

995316 

.3i 

169284 

14-55 

830716 

86 

25 

14637198923 

165454 

14-22 

995297 

•  3i 

170157 

14-53 

829843 

85 

26' 

14666   98919 

1 66307 

14-19 

995278 

-3i 

171029 

14-50 

828971 

84 

27' 

14695:98914 

167159 
168008 

i4-i6 

995260 

-3i 

171899 

14-47 

828101 

83 

2S 

14723 1 98910 

i4-i3 

995241 

•32 

172767 

14^44 

827233 

82 

29 

14732   9S906 

168856 

14-10 

993222 

•32 

173634 

14-42 

826366 

31 

80 

I478I  I9S9O2 

16970a 

14-07 

995203 

-32 

174499 

14-39 

825501 

30 
29 

81, 

14810 198897 

9-170547 

i4-o5 

9-995184 

-32 

9-175362 

14-36 

10-824638 

82 

14833  i  98893 

171389 

14-02 

995 1 65 

-32 

176224 

14-33 

823776 

28 

83 

14867  :  98889 

172230 

i3-99 

995146 

-32 

177084 

i4-3i 

822916 

27 

84 

14896 !98S84 

173070 

13-96 

993127 

•32 

177942 

14-28 

822058 

26 

85 

14925  98880 

173908 

i3-94 

995108 

•32 

178799 

14-25 

821201 

25 

86 

14954 , 98876 

174744 

i3-9i 

995089 

-32, 

179655 
i8o5o8 

14-23 

820345 

24 

87 

14932  98871 

175578 

13-88 

995070 

•32 

14-20 

819492 

23 

88 

i5oii  98867 

176411 

i3-86 

995o5i 

•32 

i8i36o 

14-17 

818640 

22 

89 

i5o4o' 98863 

177242 

13-83 

995o32 

.32 

182211 

14-15 

817789 

21 

40 
41 

10069 '98858 

178072 

i3-8o 

99501 3 

•32 

i83o59 

14-12 

816941 

20 
19 

i5o97 198854 

9.178900 

13-77 

9-994993 

^2" 

9.183907 

14-09 

10-816093 

42 

i5i26| 98849 

179726 

13-74 

994974 

-32  1 

184752 

14-07 

81524S 

18 

43 

i5i55'9S845 

i8o55i 

13-72 

994955 

•32  1 

185597 

14-04 

8i44o3 

17 

44 

i5i84;9SS4i 

181374 

13-69 

994935 

•32 

186439 

14-02 

8i356i 

16 

45 

1 52 1 2  9S836 

182196 

13-66 

9949 '6 

.33 

187280 
188120 

i3-99 

812720 
8iib8o 

15 

46 

15241 

98832 

i83oi6 

13-64 

994896 

-33 

i3-?6 

14 

47 

15270 

98827 

183834 

i3-6i 

994877 

-33 

188958 

i3-93 

811042 

13 

43 

15299 

98823 

i8465i 

13-59 

994857 

•33  i 

189794 

13-91 

810206 

12 

49 

15327 

98818 

185466 

13-56 

994838 

-331 

190629 

13-89 

8o8538 

11 

50 1 

15356 

98814 

186280 

i3-53 

994818 

•331 

191462 

13-86 

10 

61 

15385 

98809 

9-187092 

i3-5r 

9-994798 

-33 

9-192294 

i3-84 

10-807706 

"9" 

52 

i54i4 

98805 

187903 

13-48 

994779 

•33; 

193124 

i3-8i 

806876 

8 

53 

15442 

9S800 

188712 

13-46 

994759 

•  33: 

193953 

13-79 

806047 

7 

54 

1 547 1 

9S796 

lSg5ic^ 

13-43 

994739 

-33  i 

194780 

13.76 

803220 

6 

55; 

i55oo 

98791 

190323 

i3-4i 

994719 

-33, 

193606 

13-74 

804394 

5 

56 

15529 

98787 

191130 

i3-38 

994100 

-331 

196430 

13-71 

803570 

4 

57 

15557 

98782 

191933 

i3-36 

994680 

•33 

197253 

13-69 

802747 

8 

68 

15586 

98178 

192734 

i3-33 

994660 

-33' 

198074 

i3-66. 

801926 

2 

69 

15615198773 

193534 

i3.3o 

994640 

•33: 

198894 

13-64 

801106 

1 

60 

15643 j 98769 

194332 

13-28 

994620 

•33 

199713 

i3-6i 

8002S7 

0 

~\ 

N.  COS., N. sine. 

L.  COS. 

D.1" 

L.  sine. 

L.  cot. 

D.l" 

L.  tang. 

' 

81° 

TRIGONOMETRICAL   FUl^CTIOXS. — 9°. 


39 


Nat.  Functions. 

Logarithmic  Functions 

+  10. 

1 

0 

N.sineJ  N.  cos. 

L.  sine. 

D.  1" 

L.  COS. 

D.l" 

1  L.  tang. 

D.l" 

L.  oot. 

10643  98769 

9-194332 

13-28 

9.994620 

.33 

9.199713 

13.61 

10.800287 

60 

1 

n672  98764 

193129 

13-26 

994600 

.33 

1  200529 

13-59 

799471 
798655 

59 

2|J  15701  I9S760 
8^5730198755 

195925 

13-23 

994 5So 

.33 

1  201345 

13-56 

58 

196719 

l3-21 

994560 

•34 

j  202159 

13.54 

797841 

57 

4   -  -    - 

15738  98701 

197511 

i3-i8 

994540 

•34 

1  202971 

13.32 

797029 

56 

5 

15787  98746 

198302 

i3-i6 

994519 

•34 

203782 

13.49 

796218 

55 

6 

i58i6  98741 

199091 

i3.i3 

994499 

•34 

204392 

13.47 

795408 

54 

7  !i5«43l9»7^7 

199879 

i3-ii 

994479 

•34 

2o54oo 

13.45 

794600 

53 

8 

I 1D873 

92^32 

200666 

i3-o8 

994439 

•34 

206207 

13.42 

793793 

52 

9 

15902 

98728 

201431 

i3-o6 

994438 

•34 

207013 

13.40 

792987 

51 

10 
11 

1 593 1 
15959 

98723 

202234 

i3-o4 

994418 

•34 
•34 

1  207817 

13.38 

792183 

50 
49" 

98718 

9'2o3oi7 

i3-oi 

9.994397 

9.208619 

13.35 

10.791381 

12 

15988 

98714 

203797 

12-99 

994377 

•34 

1  209420 

13.33 

79o58o 

4S 

13 

16017198709 

204577 

12-96 

994357 

•34 

I  210220 

i3.3i 

789780 

47 

14 

1 16046  98704 

205354 

12-94 

994336 

•34 

:  211018 

13.28 

788982 

46 

15 

1 16074 ' 98700 

2o6i3i 

12-92 

994316 

•34 

1  2ii8i5 

13.26 

788185 

45 

16 

j i6io3  98695 
Ii6i32| 98690 
116160  98686 

206906 

12.89 

994295 

•34 

:  212611 

i3.24 

787389 
786595 

44 

17 

207679 

12.87 

994274 

•35 

i  2i34o5 

13^21 

43 

18 

208432 

12-85 

994254 

.35 

214198 

13^19 

785802 

42 

11 

16189 '98681 

209222 

12-82 

994233 

.35 

214989 

13^1? 

785011 

41 

20 

21 

16218.98676 

209992 

12-80 

994212 

J5 

213780 
9.216568 

i3^i5 

784220 

41) 
39 

16246  98671 

9-210760 

12-78 

9-994191 

.35 

13.12 

ro^8343T 

22 

16275  98667 

2II326 

12-75 

994171 

.35 

217356 

13.10 

782644 

38 

23 

i63o4  98662 

21229I 

12-73 

994 i5o 

.35 

218142 

13.08 

781858 

37 

24 

16333,  9%57 

2i3o55 

12-71 

994129 

•35 

218926 

i3.o5 

781074 

86 

25 

i636i  9S652 

2i38i8 

12-68 

994108 

.35 

219710 

13. o3 

780290 

85 

26 

163901 98648 

214579 

12-66 

994087 

.35 

220492 

i3.oi 

779508 1  34 

27 

16419  98643 

215338 

12-64 

994066 

•35 

221272 

12.99 

778728  33 

28 

16447  9'^638 

216097 

I2-6l 

994045 

•35 

222052 

12^97 

777948 

32 

29 

16476  98633 

216834 

12-59 

994024 

•35 

222830 

12-94 

777170 

31 

30 
31 

i65o5 ; 98629 
16533"  9S624 

217609 
9-218363 

12-57 

994003 

•35 

2  236o6 

12-92 

776394 

30 

12-55 

9.993981 

•35"' 

9.227382 

12^90 

10.775618  29  1 

82 

i6562  9S619 

219116 

12-53 

993960 

•35; 

225i56 

12^88 

774844 

28 

83 

16391  9S614 

219868 

I? -50 

993939 
993918 

•35! 

225929 

I2^86 

774071 

27 

34, 

16620  98609 

220618 

12-48 

•35 

226700 

12.84 

773300 

26 

85: 

16648  98604 

221367 

12.46 

993896 

•36 

227471 

I2-8l 

772529  25 

86 

16677  9^600 

222Il5 

12-44 

993875 
993854 

•36 

228239 

12.79 

771761  24 

37 

16706  98595 

222861 

12-42 

•36 

229007 

12^77 

770993  :  23 

S8 

16734  98590 
16763  98585 

223606 

12-39 

993832 

•36 

229773 

12^75 

770227 122 

39 

224349 

12-37 

993811 

•36 

23o539 

12-73 

769461  '  21  . 

40' 
"if' 

16792  98580 

223092 

12-35 

993789 

•36! 

23l302 

12.71 

768698 : 20 

16820; 98575 

9-225833 

12.33 

9.993768 

•  36 

9.232065 

12.69 

10.767935  191 

42; 

16849 ' 98570 

226573 

12-31 

993746 

•  36' 

232826 

12.67 

767174 

18 

43 

16878 1 98565 

227311 

12-28 

993725 

•36 

233586 

12.65 

766414 

17 

44  i 

16906 1 98561 

228048 

12.26 

*  993703 

.36 

234345 

12.62 

765655 

16 

f^ 

16935  98556 

228784 

12-24 

993681 

•  36 

235io3 

12-60 

764897 

15 

46 

16964  98551 

229318 

12-22 

993660 

•36 

235859 

12.58 

764141 

14 

47 

16992  98546 

230252 

12-20 

993638 

•  361 

236614 

12.56 

763386 

13 

48 

17021  1 93541 

230984 

12.18 

993616 

•36  i 

237368 

12.54 

762632 

12 

49 

i7o5o-  98536 

23i7i5 

12.16 

993594 

•37: 

238120 

12-52 

761880 

11 

r50 

l7o^8:9853^ 

232444 

12.14 

993572 

•37! 

238872 

12. 5o 

761128 

10 

17107198526 

9-233172 

12.12 

9.993550 

•37' 

9.239622 

12.48 

10.760378 

9 

52 

17136 ,98521 

233899 

12-09 

993528 

•37' 

240371 

12.46 

759629 

8 

53  1 

17164 1  98516 

234625 

12-07 

993506 

.37  i 

241118 

12.44 

758882 

7 

"i 

17193  I  9851 1 

235349 
236073 

12-05 

993484 

•37  i 

24i865 

12.42 

758i35 

6 

00  1 

17222  98506 

12-03 

993462 

•37 

242610 

12.40 

737390 

5 

K 

17250  98501 

236193 

12-01 

993440 

.37: 

243354 

12.38 

756646 

4 

57 

17279  98496 

2375i5l 

11-99 

993418 

•37 

244097 
244839 

12.36 

755903 

8 

58 

17308 >  98491 

238235 

11-97 

993396 

•37^ 

12-34 

755161 

2 

59 

17336:98486 

238953 

11.95 

993374 

•37  ■ 

243379 

12.32 

754421 

1 

60 

17365198481 

239670 

11.93 

993351 

•37 

246319 

12. 3o 

753681 

0 

N.  COS.  N.  sine. 

L.  COS. 

D.l"  1 

L.  sine. 

! 

L.  cot. 

D.l" 

L.  tang. 

80°                         1 

40 


TRIGONOMETRICAL  FUNCTIONS. — 10°. 


Nat.  Functions. 

Logarithmic  Functions 

+  10. 

1 

1 

0 

N.sine.|N.co8. 

Lsine. 

D.l" 

L.  COS.  D.l" 

L.tang. 

D.l" 

L.cot 

17365  1 98481 

9-239670 

11.93 

9.993351 

•37 

9-246819 

12.30 

10-753681 

60 

1 

17393  98476 

240886 

11-91 

993329 

•^7 

247057 

12-28 

75.2943 

59 

2 

17422  9S471 

241101 

11-89 

993307 

i'' 

247794 

12.26 

702206 

53 

3 

17451  198466 

241814 

11.87 

998283 

.37 

248380 

12-24 

751470 

57 

4 

17479198461 

242526 

11-85 

998262 

•37 

249264 

12.22 

750736 

56 

5 

17008  98455 
17337  9S450 

248287 

11-83 

998240 

:^? 

249998 

12-20 

730002 

55 

6 

248947 

ii-8i 

998195 

250780 

12.18 

749270 

54 

7 

17565:98445 

244655 

11-79 

.38 

251461 

12-17 

743389 

53 

8 

17594  93440 

245363 

11-77 
11-75 

998172 

.38 

252191 

12.13 

747809 

52 

91 

17623:98435 

246069 

993149 

•  38 

252920 

12.13 

747080 

51 

10 ! 

11 

17651  9^430 

246770 

11.73 

993127 

.381 

253648 

12.11 

746352 

50 

17680  9S425 

9-24747« 

11-71 

9.998104 

.38 

9-254874 

12.09 

10-745626 

49 

12  1 

17708  98420 

248181 

11-69 

998081 

-38 

255100 

12.07 

744900 

48 

13  1 

177^7 

98414 

248883 

11.67 

998059 

-38 

255824 

I2-05 

744176 

47 

14  { 

-17766 

98409 

249583 

11.65 

998086 

-38 

256547 

12. o3 

743453 

46 

15 

17794 

98404 

250282 

11.63 

998018 

.38 

257269 

12-01 

742731 

45 

16 

17823 

98399 

250980 

11. 61 

992990 

•  38 

257990 

12.00 

742010 

44 

17 

17852 

9S394 

251677 

n.59 

992967 

.38 

258710 

11.98 

741290 

43 

18 

17880 

fM 

252878 

11-58 

992944 

.38 

259429 

11.96 

740371 

42 

19! 

17909 

253067 

11-56 

992921 

.38 

260146 

11.94 

789834 

41 

20 : 

17987 

98878 

253761 

11-54 

992898 

.38 

260868 

11.92 

739187 

40 

21' 

17966 

98873 

9-254453 

11.52 

9.992875 

.38  i 

9.261578 

11.90 

10-788422 

3i) 

22 

17993 

98868 

255x44 

11. 5o 

992852 

•  38 

262292 

11.89 

737708 

33 

23 

18023 

98862 

255834 

11.48 

992829 
992806 

.39 

263oo5 

11.87 

736993 
786288 

37 

24 

i8o52 

98357 

256523 

11.46 

.39 

268717 

11-85 

86 

25 

1 808 1 

93352 

257211 

11-44 

992788 

.39 

264428 

11-83 

735572 

35 

26 

18109 

98847 

257898 

11.4s 

992759 

•39 

265i38 

ii-8i 

734S62 

34 

27 

i8i38 

98341 

2585b3 

II. 41 

992786 

.39 

265847 

11-79 

734153 

33 

28 

18166 

9S336 

259268 

11-89 

992713 

•39, 

266555 

11.78 

788445 

32 

29 

18195 

98331 

259951 

11.37 

992690 

.39 

267261 

11.76 

782739 

31 

30 
31 

18224 

98825 

260633 

11.35 

992666 

.39 

267967 

11.74 

782033 

30 

18252 

98820 

9-26i3i4 

11.33 

9-992643 

•39. 

9.268671 

11-72 

10-781829 

29 

82 

18281 

98815 

261994 

II. 3i 

992619 

•39, 

269875 

11-70 

780625 

23 

33 

i83o9 

98310 

262673 

II. 3o 

992596 

•39, 

270077 

11.69 

729928 

27 

34 

18338 

98304 

263351 

11-28 

992372 

.39, 

270779 

11.67 

729221 

26 

35 

18867 

98299 

264027 

11.26 

992349 

.39 

271479 

11-63 

728521 

25 

36 

18895 

98294 

264703 

11.24 

992323 

.39 

272178 

11.64 

727822 

24 

37 

18424 

98268 

265377 

11.22 

992301 

.39 

272876 

11-62 

727124 

23 

38 

18452 

98288 

266o5i 

11.20 

992478 

•40 

273573 

11.60 

726427 

22 

39 

1 848 1 

98277 

266723 

11.19 

992454 

.40 

274269 

11.58 

723781 

21 

40 
41 

i8509 

98272 

267895 

II. 17 

992480 

.40 

274964 

11.57 

725o36 

20 

i8538 1 98267 

9-26do65 

11. i5 

9-992406 

.40 

9-275658 

11-55 

10.724342 

ly 

42 

18567 

98261 

268734 

iii3 

992882 

.40 

276351 

11-53 

728649 

18 

43 

18595 

98256 

269402 

ii'ii 

992839 

-40 

277043 

n-5i 

722957 

17 

44 

18624 

98250 

270069 

II. 10 

992335 

.40 

277734 
278424 

11-50 

722266 

16 

45 

i8652 

98245 

270735 

11-08 

992811 

•  40 

11-48 

721576 

15 

46 

18681 {98240 

271400 

11-06 

992287 

-40 

279113 

11.47 

720S87 

14 

47 

18710 198234 

272064 

ii-o5 

992268 

.40 

279801 
280488 

11.45 

720199 

13 

48 

18738:98229 

272726 

11-03 

992289 

-40 

11.43 

710012 
718S26 

12 

49 

18767:98228 

278888 

11. 01 

992214 

-40 

281174 

II. 41 

11 

50 
51 

18795 j 98218 

274049 

10.99 

992190 

-40 
•40 

28i858 

11.40 

718142 

10 

18824  98212 

9-274708 

10.98 

9.992166 

9.282542 

11-38 

10-717458 

9 

52 

i8852  98207 

275367 

10.96 

902142 

•40 

283^25 

11-36 

716775 

8 

53 

18881  98201 

276024 

10.94 

992II7 

-41 

288907 
284388 

11-35 

716093 

7 

54 

18910  98196 

276681 

10.92 

992098 

•41 

11-33 

715412 

6 

55 

18988  98190 
18967  98185 

277337 

10.91 
10.89 

992069 

-41 

285268 

ii-3i 

714732 

5 

56 

277991 

992044 

-41 

285947 

11-80 

714053 

4 

57 

18995  98179 

278644 

10.87 

992020 

-41 

286624 

11.28 

713376 

8 

58 

19024  98174 

279297 

10.86 

991996 

•41 

287801 

11-26 

713699 

2 

59 

19052  98168 

279948 

10.84 

99I97I 

-41 

287977 

11-25 

712028 

1 

60 

19081  98163 

2S0599 

10.82 

991947 

-41 

288652 

11-23 

711848 

0 

N.  COS.  N.  sine. 

L.  COS. 

D.l" 

L.  sine. 

L.  cot 

D.l" 

L.  tang. 

' 

T9°                        1 

trigox(5metkical  junctions. — 11= 


41 


42 


TRIGONOMETRICAL   FUXCTIOXS.— 12= 


Nat.  Functions. 

Logarithmic  Functions  +  10.            1 

~0 

N.  sine  J  N.  cos. 

L.  sine. 

D.  1" 

L.  COS.   D.l" 

L.  tang.  1  D.  1" 

L.  cot. 

20791  '   97815 

9-317879 

ITs 

9-990404  -45 

9-327474!  10-35 

10-672526 

60 

1 

20820  97809 

318473 

990378  -45 

328095!  10-33 

671905 

59 

2 

20848  !  97803 

319066 

9-87 

99o35i  -45 

328715 
329334 

10-32 

671285 

58 

3 

20877  ;  97797 

319658 

9-86 

990324-45 

io-3o 

670666 

57 

4i  2090D  977QI 

320249 

9.84 

990297! .45 

329953-  10-29 
330570  10-28 

670047 

56 

5 

209JJ  97784 

320840 

9-83 

990270 

•45 

669430 
6688. 3 

55 

6 

20962  97778 

32i43o 

9-82 

990243 

•45 

331187  10-26 

54 

7 

20990  97772 

322019 

9.80 

9902x5 

-45 

331803;  10-25 

668197 

53 

8 

21019!  97766 

322607 

9-79 

990188 

•45 

332418'  10-24 

667582 

52 

9 

21047;  97760 

323194 

9-77 

990161 

•45 

333o33'  10-23 

666967 
666354 

51 

10 
11 

21076:  97754 

323780 

9-76 

990134 

•45 

333646;  10-21 

50 
49 

21104  97748 

'g^Uibb 

9-75 

9-990107 

.46 

9-334259]  10-20 

10-665741 

12 

2Il32   97742 

324950 

9-73 

990079 

.46 

334S71  1  10-19 

665129 

43 

13 

2II61  !  97735 

325534 

9.72 

990052 

.46 

335482'  10-17 

664518 

47 

14 

2II89!  97729 

326117 

990025 

.46 

3360931  10-16 

663907 

46 

15 

21218!  97723 

326700 

9-69 

989997 

■46 

336702  10- i5 

663298 

45 

16 

21246  1  97717 

327281 

9-68 

9S9970 

.46 

337311I  10-13 

662689 

44 

17 

21273  1  97711 

327862 

9-66 

989942 

.46 

337919 

10-12 

662081 

43 

18 

2i3o3  1  97705 

328442 

9-65 

9899.5 

-46 

338527 
339133 

lO-II 

661473 

42 

19 

2i33i  1  97698 

329021 

9-64 

989887 

.46 

10- 10 

660867 

41 

20 
'2! 

. 2i36o 

97692 

329599 

9-62 

989860 

•46 

339739 

io-o8 

660261 

40 

'21388 

976S6 

9-330176 

9-6i 

9-989832 

-46 

9-340344 

10-07 

10-659656 

39 

2i. 

21417 

97680 

330753 

9 -60 

9^9804 

.46 

340948 

10-06 

659o52 

38 

23 

21445 

97673 

33i329 

9-58 

9^9777 

.46 

341552 

10-04 

658448 

37 

24 

21474 

97667 

331903 

9-57 

9S9749 

•47 

342.56 

io-o3 

657845 

36 

25 

2l5o2 

9766. 

332478 

9-56 

989721 

•47 

342757 

10-02 

657243 

35 

26 

2i53o 

97655 

333o5i 

9-54 

989693 

-47  1 

343358 

10-00 

656642 

34 

27 

2.559 

97648 

333624 

9-53 

9S9665 

•47 

343938 

9-99 

656042 

33 

2S 

2087 

97642 

334195 

9-52 

9S9637 

•47 

344558 

9-98 

655442 

32 

29 

2I6I6 

97636 

334766 

9 -So 

989609 

•47  1 

345 1 57 

9-97 

654843 

81 

SO 

21644 

97630 

335337 

9-49 

989582 

•47  1 

343755 
9-346353 

9-96 
9-94 

654245 
.0-653647 

30 
"29" 

21672 

97623 

9-335906 

9.48 

9-989553 

•47, 

32 

! 21701 

97617 

336475 

9.46 

989525 

■47. 

346949 

9-93 

653o5i 

23 

33 

: 21729 

97611 

337043 

9.45 

989497 

•47 

347343 

9-92 

652455 

27 

34 

'21758 

97604 

337610 

9.44 

9^9469 

•47! 

348141 

9-91 

65.859 

26 

35 

21786!  97598 

338176 

9-43 

98^441 

-47 

348735 

9.90 

661265 

25 

36 

21814I  97392 

338742 

9-41 

9S9413 

•47 

349329 

9-88 

650671 

24 

37 

! 21843  97585 

339306 

9.40 

989384 

•47 

349922 
33o5i4 

9-87 

650078 

23 

38 

: 21871  97579 
218991  97573 

339871 

9-39 

9S9356 

•47 

9-86 

649486 

22 

39 

340434 

9.37 

989328 

•47; 

"35iio6 

9-85 

648894 

21 

40 
"if 
42 
43 

■  219281  97566 

340996 

9-36 

989300 

•47  1 

35.697 
9-352287 

9-83 

6483o3 

20 

21956  97560 

9-34i558 

9.35 

9-989271 

•47; 

9-82 

10-647713 

19 

^2.985  97553 

342119 

9-34 

989243 

•47 

352876 

9-8. 

647124 

18 

22013 i  97547 

342679 

9.32 

9892.4 

•47 

353465 

9-80 

646535 

17 

44 

22041 ;  97541 

343239 

9-3i 

989.86 

•47 

354053 

9-79 

645947 

16 

45 
46 

47 

220701  97534 

343797 

9 -30 

989.57 

•47 

354640 

9-77 

645360 

15 

22098  97528 

344355 

9-29 

989128 

.48 

355227 

9.76 

644773 

14 

22126  97521 

344912 

9-27 

989100 

.48 

3558.3 

9.73 

644187 

13 

48 

221551  975i5 

345469 

9-26 

989071 

.48 

356398 

9-74 

643602 

12 

49 

22i83;  97508 

346024 

9-25 

989042 

•481 

356982 
357566 

9.73 

643oi8 

11 

50 
"5]" 
52 
58 

22212 

97502 

346579 

9-24 

9890.4 

•481 

9-71 

642434 

10 

22240 

97496 
97489 

9-347134 

9-22 

9-988985 

•48: 

9-358149 

9-70 

io-64i85i 

9 

22268 

347687 

9.21 

988956 

•48  1 

358731 

9-69 
9-68 

641269 

8 

22297 

97483 

348240 

9-20 

988927 

-48 

359313 

640687 

7 

54 

223251 

97476 

348792 

9-19 

988898 

•48! 

359893 

9-67 

640107 

6 

55 

22353] 

97470 

349343 

9-17 

988869  -48! 

360474 

9-66 

639526 

5 

56 

22382 

97463 

349S93 

9-i6 

988840 

•48: 

36io53 

9-65 

638947 

4 

57 

22410 

97457 

350443 

9. ,5 

98881 1 

•49' 

36i632 

9-63 

638368 

3 

5S 

22438! 

9745o 

350Q92 

9-14 

988782 

•49; 

362210 

9-62 

637790 

2 

59 

2246-:  i 

97444 

33ID40 

9-i3 

988753 

•49  1 

362787!  9-6i 

637213 

1 

60  , 

22495 

97437 

352088 

9. II 

988724  -49 

363364   9 -60 

636636 

0 

n'coT: 

J?,  sine. 

L.  COS. 

D.l"  1 

L.  sine.  1    i  L.  cot.  !  D.  1"  i 

L.  tang. 

~ 

77^^                        1 

TRIGONOMETRICAL    FUNCTIONS. — 13° 


43 


Nat.  Functions. 

LoiiARiTHMic  Functions 

^-  10. 

1 

0  ■ 

N.sine.  N.  cos. 

L.  sine. 

D.  1" 

L.  COS. 

D.l" 

L.  tang. 

Dl."  j 

L.  cot 

22495  97437 

9-352088 

9-11 

9-988724 

•49 

9 •363364 

9-60 

10. 636686 

60 

1 

22523  97430 

352635  j 

9-10 

988695 

.49 

363940 

9.59 

686060 

59 

2j! 

22552 i 97424 

353i8i 

9.09 

988666 

.49 

36451 5 

9-58 

635485 

53 

8 

2258o  97417 

353726 

9-08 

988636 
988607 

•49 

365090 

9-57 

684910 

57 

4!' 

22608  9741 1 

354271 

9-07 

•49 

365664 

9-55 

634336 

56 

5 

22637  97404 

354^:115 

9-o5 

988578 

•49 

366237 

9-54 

688763 

55 

6 

22665 19-398 

3553581 

9.04 

988548 

•49 

366810 

9-58 

688190 

54 

7 
8 
9 

22693 

97391 

355901 1 

9 -03 

988519 

.49 

367382 

9-52 

682618 

53 

22722 

97384 

356443 

9-02 

988489 

•49 

367953 

9-51 

682047 

52 

22750 

97378 

356934 1 

9-01 

938460 

•49 

368524 

9-5o 

681476 

51 

10 ! 

^778 

97371 

3070241 

8-99 

988430 

•49 

369094 

9-49 

680906 

50 

iM 

22807 

97365 

9-353064 

8-98 

9-988401 

•49 

9^369663 

9-48 

io.63o337 

49 

12 

22835^97358 

3586o3 

8-97 

988371 

•49  1 

870232 

9-46 

629768 

43 

131 

22863! 97351 

359141 

8-96 

988342 

•49; 

370799 

9-45 

629201 

47 

ui 

22892:97345 

359678 

8-95 

988312 

-So! 

371367 

9-44 

628633 

46 

15  i 

22920197338 

36021 5 

8-93 

988282 

-So! 

371933 

9-43 

628067 

45 

16 

22948  97331 

360752 

8-92 

988252 

-50  1 

372499 

9-42 

627001 

44 

17 

22977  97325 
23oo5  97318 

361287 

8-91 

988223 

-5o! 

373064 

9-41 

626986 

43 

13 

361822 

8.90 

988193 

.50 

373629 
374193 

9-40 

6263TI 

42 

19 

23o33|973ii 

362356 

8-89 
8-88 

988163 

-So 

lit 

623807 

41 

2u 

21 
!  22 

23o62  97304 

362889 

988133 

-5o 

374756 

623244 

40 

23090 

97298 

9-3634?2 

8-87 
8-85 

9-988103 

-So, 

9^3753i9 

9-37 

10-624681 

89 

23ii8 

97291 

363934 

988073 

-5o 

375881 

9.35 

624119 

33 

2:} 

23 146 

97284 

364485 

8-84 

988043 

-So 

376442 

9-34 

623558 

37 

21 

28175 

97278 

365oi6 

8-83 

988013 

-So 

377003 

9.33 

622997 

36 

25 

23203 

97271 

365546 

8-82 

987983 

-So 

377563 

9-32 

622487 

85 

20 

2323l 

97264 

366075 

8-81 

987953 

-So 

878122 

9-3i 

621878 

34 

27 

23260 

97257 

366604 

8-80 

lt]tll 

-5o 

878681 

9-80 

621819 

33 

2S 

23288 

97251 

367i3i 

8-79 

-50 

879289 

9-20 

9-28 

620761 

82 

29 

233i6 

97244 

367639 

8-77 

987862 

.50 

379797 

620208 

31 

80 
81 

23345 

97237 

368i85 

8-76 

987832 

-51 

38o354 
9-880910 

9.27 
9-26 

61-9646 

10-619090 

618534 

80 

29 

23373 

97230 

9-368711 

8-75 

9-987801 

.5i 

32 

23401 

97223 

369236 

8-74 

987771 

-5i 

881466 

9-25 

23 

88 

23429 

97217 

369761 

8-73 

987740 

-5i 

882020 

9-24 

617980 

27 

84 

23458 

97210 

370285 

8-72 

987710 

.5i 

882575 

9-23 

617425 

26 

85 

23486 

97203 

370808 

8-71 

987679 

.5i 

888129 

9-22 

616871 

25 

8(5 

235i4 

97196 

371330 

8.70 

987649 

-5i 

383682 

9^21 

616818 

24 

87 

23542 

97189 

371852 

8-69 

987618 

-51 

884284 

9^20 

615766 

23 

83 

23571 

97182 

372373 

8-67 
8-66 

98^588 

.5x 

884786 

9.19 

6i52i4 

22 

89 

23599 

97176 

372894 

987557 

-5i 

885337 

9-ia 

614668 

21 

40 
41 

23627 

97169 

373414 

8-65 

987526 

.5. 

385888 

9-17 

614112 

20 

23656 

97162 

9-373933 

8-64 

9-987496 

-5i 

9.886488 

9^i5 

io^6i3l>b2 

TiT 

42 

23684 

97155 

374452 

8-63 

987465 

-5i 

886987 

9-14 

618018 

13 

43 

23712 

97148 

374970 

8-62 

987434 

-51 

887536 

9-i3 

612464 

17 

44 

23740 

97141 

373487 

8-61 

987403 

-52 

888084 

9-12 

611916 

16 

45 

23769 

97134 

376003 

8-60 

987372 

-52 

888631 

9-u 

611869 

15 

46 

23797 

97127 

376519 

8-59 
8-58 

987341 

•52 

889178 

9-10 

610822 

14 

47 

23825 

97120 

377033 

987310 

-52 

889724 

9-00 
9-08 

610276 

13 

43 

23853 

97113 

377549 

8-57 

987279 

•52 

890270 

609780 

12 

4y 

23882 

97106 

378063 

8-56 

987248 

•52 

890815 

9-07 

609185 

11 

•0 
51 

23910 

97100 

378577 

8-54 

987217 

•52 

891860 

9 -06 

608640 

10 

23938 

97093 

9-379089 

8-53 

9-987186 

•52 

9-891908 

9 -05 

10-608097 

"9" 

52 

23966 

97086 

3-9601 

8-52 

987155 

•52 

892447 

9.04 

607533 

8 

53 

1  23995 

97079 

3Soii3 

8-51 

987124 

-52 

892989 

9-08 

607011 

7 

54 

24023 

97072 

330624 

8-5o 

987092 

•52 

893531 

9-02 

606469 

6 

55 

24o5i 

97063 

38ii34 

8-49 

987061 

•52 

894078 

9-01 

603927 
6o5386 

5 

56 

24079 

970D3 

3Si643 

8-48 

987030 

-52 

894614 

9-00 

h 

4 

57 

24108 

97o5i 

382152 

8-47 

986998 

-52 

395154 

604846 

3 

58 

24i36 

97044 

332661 

8-46 

986967 

-52 

395694 

604306 

2 

59 

124164:97037 

333168 

8-45 

986936 

-52 

896233 

8-97 

608767 

1 

60 ;!  24192  |97o3o 

333675 

8-44 

986904 

•52 

1  396771 

8-96 

608229 

0 

IjN.  COS.  N.  sine. 

j  L.  COS. 

I  D.l" 

L.  sine. 

1  L.  cot. 

D.l" 

L.  tang. 

' 

76<^                         1 

44 


TRIGOXOMETRIC AL   F CXCTIOXS. — 14". 


Nat.  Functions. 

Logarithmic  Fuxctioxs  +  10.             1 

0 

N.sine.  X.  cos. 

L.  sine. 

D.l" 

L.  COS.   D.l   L.  tang.   D.  1" 

L.cot  1 

24192  97o3o 

9.353675 

8.44 

9.986904  .52  9-396771 

8.96 

io-6o3229  60 

1 

24220  9-023 

3S4182 

8-43 

986873 

.53  ,  397309 

896 

602691 

59  1 

2 

24249  QTOl5 

384687 

8.42 

986841 

•53 

397846 

8.95 

602154  53  1 

8  24277  9700S 

385i92 

8-41 

986809 

.53 

.  398333 

8-94 

601617  57  1 

4  243o5  Q'ooi 

385697 

8.40 

9S6778 

.53 

398919 

8-93 

601081 

56 

/  5 

24333  96994 

386201 

8. "3^ 

9S6746 

.53 

399455 

8-92 

600545 

55 

6 

24362  96987 

3S6704 

95^714 

.53 

399990 
400524 

8-91 

600010 

54 

7 

24390  969S0 

3S7207 

^^7 

986683 

•53 

8-90 

599476 

53 

8 

24418  96973 

387-09 

8-36 

986631 

•53 

401 o53 

8-89 

598942 

52 

9 

24446  96966 

38^210 

8.35 

986619 

•53 

401591 

8-88 

598409 

51 

10 

24474  96959 

388711 

8.34 

986587 

•53 

402124 
9-402656 

8.87 

597876  50 

11  245o3  q6q52 

9-389211 

8-33 

Q. 986555 

•53 

8-86 

10.597344149 

12 

2j53i  96945 

3S9711 

8.32 

986523 

.53 

4o3i87 
40371a 

8.85 

5o68i3  43 

13 

24559  96937 

390210 

8-31 

986491 

.53 

8.84 

596282  47 

14 

245S7  96930 
24615  96923 

390708 

8-30 

986459 

.53 

404249 
^  404778 

8.83 

595751  46 

15 

391206 

8.28 

986427 

•53 

8-82 

595222  45 

16 

24644  96916 

391703 

8.27 

986395 

•53 

4o53o8 

8-81 

594692 1 44 

1^, 

24672 ' 96909 

'& 

8.26 

986363 

•54 

405836 

8-80 

594 1 64  1 45 

18; 

24700  96902 

8-25 

986331 

.54 

406364 

8.70 
8.78 

593636 1 42 

19' 

24728  96894 

393191 

8.24 

986299 

.54 

406892 

593108  41 

20; 
21; 

24756 ' 96SS7 

393685 

8-23 

986266 

•54 

407419 
9.407945 

8.77 

592581  40 

24784 ■ 96880 

9.394179 
394673 

8.22 

9.986234 

.54 

8.76 

10-592055! 89 

221 

24813.96S73 

8.21 

986202 

•54 

,  408471 

8.75 

591529133 
591003  37 

23  i 

24841  1  96866 

395166 

8-20 

986169 

.54 

408997 
409321 

8.74 

24 

24869 1 96S58 

395658 

8-19 
8.18 
8.17 

986137 

•54 

8.74 

590479 

36 

25 
26 

24897  1 9685 1 
24925 1 96844 

396100 
396641 

986104 
986072 

•54 
•54 

410045 
410569 

8.73 
8.72 

589955 
589431 

35 
34 

27 

24954 1 96837 

397132 

8.17 

986039 

•54 

41 1092 

8.71 

588908 

33 

28 

24982196829 
25oio 196822 
25o38| 96815 

39-621 

8.16 

986007 

•54 

4ii6i5 

8.70 

588385 

32 

29 

398111 

8.i5 

985974 

•54 

412137 

8.69 
8.68 

587864 

31 

30 
31, 

39^600 

8.14 

985942 

•54 
.55 

412658 
9.413179 

587342 

30 

25o66 :  96807 

"9^99088 

8-13 

''$^?6 

8-67 

10.586821 

29 

S21 

25094 1 96800 

399575 

8.12 

.55 

413699 

8-66 

586301 

23 

S3' 

25l22  96793 

400062 

8.11 

985843 

.55 

414219 

8-65 

535781 

27 

84; 

25i5i  96-S6 

400549 

8-10 

985811 

.55 

414735 

8-64 

585262 

26 

85' 

25179:96778 

4oio35 

8.09 

985778 

.55 

415257 

8.64 

584743 

25 

86' 

25207  07"! 

401320 

8-08 

985745 

•  55 

415775 

8.63 

584225 

24 

87 

25235  96-64 

402005 

8-07 

985712 

•  55 

416293 

8.62 

583707 

23 

8S 

25263  96706 

402489 

8.06 

985679 

.55 

416810 

8.61 

583190 

22 

39 

25291  96749 

402972 

8-05 

985646 

.55 

417326 

8-6o 

582674 

21 

40 
41 

25320  96-42 

403455 

8-04 

98561 3 

•  55 

417842 

8.59 

582158 

20 

25348  96734 

9.403933 

8-o3 

9.985580  1 

.55 

9.418353 

8-58 

10.581642 

19 

42 

25376  96-27 

404420 

8-02 

985547  1 

.55 

j  418873 

8-57 

581127 

13 

43 

25404  96719 

404901 

8-01 

985514 

.55 

419337 

8-56 

58o6i3 

17 

44 

25432  96712 

4o5382 

8.00 

985480 

.55 

419901 

8-55 

p?? 

16 

45 

25460 ' 96705 

4o5S62 

]-ll 

985447 

.55 

42o4i5 

8-55 

15 

46' 

25488 , 96697 

406341 

985414 

•  56 

420927 

8-54 

l]r£ 

14 

47 

2  55 1 6  96690 
25545 : 96682 

406820 

7-97 

985380 

•  56 

421440 

8-53 

13 

43 

407299 

7.96 

985347 

•  56 

421952 

8-52 

578048 

12 

49 

25573 ! 96675 

407777 

7.95 

985314 

•  56 

422463 

8-5i 

577537 

11 

50 
51 

25601 '96667 

408254 

7-94 

985280 

•  56 

422974 

8-5o 

577026 

10 

25629 '  96660 

9.408731 

7-94 

9.985247  i 

•  56 

9.423484 

8-40 
8-48 

10.576316 

9 

52 

25657 , 96653 

409207 

7.93 

985213  1 

•  56 

423993 
4245o3 

576007 

8 

53 

25685  96645 

409682 

7.92 

985 180 

•  56 

8-48 

f^in 

7 

54 

25713  96633 

410157 

7.91 

985146 

•  56 

425oii 

8-47 

6 

55 

25741  96630 

410632 

985,13 

•  56 

425519 

8-46 

574481 

5 

56 

25760 ■ 06623 

411106 

985079  1 

•  56 

426027 

8-45 

573973 

4 

57  257qS  066 1 5 

41 1 579 

985045  ; 

•  561:  426534 

8-44 

573466 

8 

58 

25826  96608 

4l2o52 

7-87 

9S5011 

•  56 1:  427041 

8-43 

572939 

2 

59 

25854  96600 

412524 

7.86 

984978 

•56  |i  427547 

8.43 

572453 

1 

60 

25882  96593 

412996 

7-85 

984944 

•561  428052 

8.42 

571948 

0 

N.  COS.  N.  sine. 

L.  COS. 

1  D.l" 

L.  sine.  1    1  L.  cot. 

D.l" 

L.  tang. 

» 

75° 

1 

TRIGONOMETRICAL  FUNCTIONS. — 15°. 


45 


Nat.  Functions. 

LOGARiTHiiic  Functions 

+  10. 

1 

M 

N.  sine.  N.  cos. 

L.  sine. 

D.l" 

1  L,  cos.  |D.l":j  L.  tang. 

D.l" 

L.cot. 

oi 

25832 '06393 

9-412996 

7-85 

9.984944 

.57 19.428052 

8.42 

10-571948 

60 

1; 

25910 :965a5 

413467 

7-84 

984910 

•57' 

428007 

8.41 

571443 

59 

2 

25988  96578 

418938 

7-83 

984876 

•57 

429062 

8-40 

670988 

5S 

8 

25966  96370 

414408 

7-83 

984842 

•57 

429566 

8.39 

570434 

57 

4 

25994  96062 

414878 

7-82 

984808 

•57 

480070 

8.38 

569980  1  5(5 

5 

26022  96555 

415347 

7-8i 

984774 

•57 

480573 

8.38 

569427  j  55 
568920  j  54 

6 

26o5o 

96D47 

4i58i5 

7-80 

984740 

•57 

481075 

8.37 

7 

26079 

96040 

416283 

7-79 

984706 

•57 

43 1 577 

8.36 

568423  53 

8 

26107 
26135 

96032 

416751 

7-78 

984672 

•57 

482079 

8.35 

567921  52 

9 

96524 

417217 

7-77 

984687 

•57 

432580 

8.34 

•  567420  1  51 

10 
11 

26163 
26191 

965n_ 

417684 

7-76 

984603 

•57 

433o8o 

8.33 

566920 

50 

96509 

9-4i8i5o 

7-75 

9.984569 

•57 

9.433580 

8.32 

10-566420 

4 'J 

12 

26219 

96502 

4i86i5 

7-74 

984535 

•57 

434080 

8.32 

565920 

43 

13 

26247  i  96494 

419079 

7-73 

984000 

•57 

434579 

8.3i 

565421 

47 

14 

26275  I  96486 

419544 

7-73 

984466 

•57 

435078 

8.3o 

564922 

46 

15 

263o3  i  96479 

420007 

7-72 

984432 

.58 

435576 

8.29 

564424 

45 

16 

26331 1 96471 

420470 

7.71 

984397 

.58 

486073 

8.28 

568927 

44 

17 

26359 ; 96463 

420933 

7-70 

984868 

.58 

436570 

8.28 

563430  i  4:5  1 

18 

263S7 1 96456 

421390 

]:ll 

984828 

.58 

Sltl 

8.27 

562983 

42 

19 

26415:96448 

421857 
4223i8 

984294 
984269 

.58 

8-26 

562487 

41 

20 
21 

26443  I  96440 

7-67 

.58 

488059 

8-25 

561941 

40 

26471 196433 

9.422778 

7.67 

9.984224 

.58 

9.438554 

8.24 

10-561446 

39 

22 

26500:96425 

423238 

7.66 

984190 

.58 

489048 

8.23 

560952 

33 

23 

26528  96417 

428697 

7-65 

984155 

.58 

489543 

8-23 

660457 

37 

24 

265561 96410 

424 1 56 

7-64 

984120 

.58 

440086 

8.22 

559964 

3G 

25 

26584 

96402 

424615 

7-63 

984085 

.58 

440529 

8.21 

609471 

35 

26 

26612 

96394 

425073 

7-62 

984050 

.58 

441022 

8.20 

608978 

34 

27 

26640 

96386 

425530 

7-6i 

984015 

.58 

44i5i4 

8.19 

558486 

33 

28 

26668 

96379 

425987 

7 -60 

988981 

•  58 

442006 

8.19 

607994 

32 

29 

26696 

96371 

426443 

7-6o 

988946 

.58 

442497 
442988 

8.18 

557508 

31 

80 

26724 

96363 

426S99 

7-59 

988911 

.58 

8.17 

667012 

30 

81 

26752 

96355 

9-427354 

7-58 

9.988875 

.58 

9.443479 

8.16 

IO-65602I 

2ir 

82 

26780 

96347 

427809 
428263 

7-57 

988840 

.59 

1  443968 

8.16 

606082 

28 

83 

26808 

96340 

7-56 

988800 

.59 

1  444458 

8.i5 

500042 

27 

84 

26836 

96332 

428717 

7-55 

988770 

.59 

444947 

8-14 

55oo53 

26 

85 

26S64 

96324 

429170 

7-54 

988735 

.59 

445435 

8.i3 

654565 

25 

86 

26892 

96316 

429623 

7-53 

988700 

.59 

445923 

8.12 

664077 
553589 

24 

87 

26920 

96308 

480075 

7-52 

988664 

.59 

44641 1 

8.12 

23 

88 

26948 

96801 

480527 

7-02 

988629 

.59 

446898 

8. II 

658102 

22 

89 

26976 

96293 

480978 

7-5i 

9S3594 

.59 

447384 

8.10 

662616 

21 

40 
"if 

27004 

96285 

481429 

7-5o 

983558 

.59 

447870 

8-09 

552 1 3o  20 

27032 

96277 

9.431879 

7-49 

9.983523 

.59 

9-448356 

8-09 

10.551644 |iy 

42 

27060 

96269 

482829 
432778 

]■■% 

988487 

.59 

448841 

8.08 

551169  13 

43 

27088 

96261 

983452 

.59 

449326 

8-07 

600674 ' 17 

44 

127116196253 

488226 

I'^l 

983416 

.59 

449810 

8-06 

600190 !  l() 

45 

127144  96246 

488675 

7-46 

988881 

.59 

450294 

8.06 

6^9706 ! 15 

46 

27172  96238 

434122 

7-45 

988845 

.59 

450777 

8.o5 

649228  u 

47 

27200  96230 

434569 

7-44 

& 

•So 

451260 

8-04 

648740  13 

48 

27228 j 96222 

485oi6 

7-44 

.60 

451743 

8-o3 

648267  12 

49 

1272561 96214 

435462 

7-43 

988233 

.60 

452225 

8.02 

647775  u 

51 

j  272-84 '  96206 

480908 

7-42 

988202 

.60 

402706 

8-02 

647294  i 10 

; 27312196198 

9.486853 

7-41 

9.988166 

.60 

9.453187 

458668 

8-01 

10-546813 

9 

52 

27340196190 

486798 

7-40 

988130 

.60 

8-00 

646882 

8 

58 

27368 

96182 

437242 

7-40 

988094 

.60 

454148 

7-99 

546862 

7 

54 

27396 

96174 

437686 

]-M 

983o58 

.60 

454628 

7-99 

645372 

6 

55 

27424 

96166 

488129 

988022 

.60 

455107 

7.98 

644898 

5 

56 

27452 

96158 

488572 

1-^1 

982986 

-60 

455586 

7-97 

644414 

4 

67 

274^0 

96i5o 

439014 

7.36 

982950 

-60 

406064 

7.96 

643986 

3 

58 

27008 

96142 

439456 

7-36 

982914 

.60 

456542 

7.96 

643458 

2 

59 

27536 

96134 

439897 
440338 

7.35 

982878 

-60 

457019 

7.95 

542981 

1 

60; 

27564  96126 

7-34 

982842 

.60 

457496 

7-94 

642604 

0 

~| 

N.  COS.  N.  sine.  { 

L.  cog. 

D.l" 

L.  sine. 

L.cot. 

D.l" 

L.  tang,  j  '  1 

I 

46 


TRIGONOMETRICAL   FUNCTIONS.— 16' 


Nat.  FiTKCTiONs. 

LOGABITHMIC  FUNCTIONS 

+  10. 

' 

N.  sine.  N.  cos. 

Ksine.  j 

D.l" 

L.  COS.   D.l"  L.  tang. 

D.l" 

L.  coL 

Oi 

27364  96126 

9-440338; 

7-34 

9.982842  1  -60  9.437496 

7-94 

lo- 042504 

60 

1 }  27092  961 18 

440778 

7.33 

982800 

.60 

437973 

7-93 

542027 

59 

2,  27620  96110 

441218: 

7.32 

982769 

.61 

438449 

7-93 

54i55i 

53 

3  2764S  9'Jio2 

441658  J 

7-3. 

982733 

•61 

458925 

7-92 

541075 

57 

4  27676  9O094 

442096 ; 

7-3i 

982696 

•61 

439400 

7.91 

040600 

06 

5  27704  96086 

442535  ; 

7-30 

982660 

•61 

459873 

7-90 

540125 

55 

6 j  27731  96078 

442973 ' 

7-29 

982624 

.61 

460349 

7-90 

539651 

54 

7  j'  27759  96070 

443410 

7-28 

982087 

•61 

460823 

7.89 

539177 

53 

8  127787  96062 

443847 • 

7.27 

982031 

.61 

461297 

7^88 

538703 

52 

9'  278i5  q6o54 

444284 

7.27 

982514 

.6. 

461770 

7.88 

538230 

51 

10' 

27S43  96046 

444720 

7-26 

982477 

•  61  1 

462242 

7-87 

537753 

50 
49 

11 

27871  96037 

9.443155 

7-20 

9.982441 

.61  ^62714 

7.86 

10-537286 

12 

27899  96029 

445590 

7-24 

982404 

•61 

463186 

7-?^ 

536814 

43 

13 

27927  96021 

446025 

7-23 

982367 

.61 

463658 

7-80 

536342 

47 

14 

27955  96013 

446439 

446893 

7-23 

982331 

.61 

464129 

7-84 

535871 

46 

15 

27983  96005 

7.22 

982294 

.61 

464599 

^1^ 

535401 

45 

16 

28011  95997 

447326 

7.21 

982237 

.61 

465069 

7-83 

534931 

44 

17 

28039  9^939 

447739 

7.20 

982220 

.62 

465539 

7-82 

534461 

43 

13 

28067  93981 

443191 

7.20 

982183 

.62 

466008 

7.81 

533992 

42 

19 

28oq5  q5972 

443623 

7.19 

982146 

.62 

466476 

7.80 

533524 

41 

20  28123  93964 

449034 

7-18 

982109 

.62  1 

466945 

7.80 

533o55 

40 

21  .  2Si5i  95950 

9.449435 

7.16 

9.932072 

.62  0.467413 

7-79 

10-532587 

39 

22  28178  9594S 

449910 

982035 

.62  1 

467880 

7.78 

532 1 20 

33 

23  28206  93940 

45o345 

7.16 

981998 

.621 

468347 

7.78 

53i653 

37 

24  28234  93931 

450770 

7.10 

981961 

.62 

468814 

7-77 

531186 

30 

25  28262  95923 

451204 

7-U 

981924 
981886 

.62 

469280 

7.76 

530720 

35 

26  ' 28290  95913 

45i632 

7-13 

.62 

469746 

7.75 

530254 

34 

27  ;  283 18  95907 

452060 

7-i3 

981849 

•02 

470211 

7.75 

029789 

33 

28 ;  28346-  93S98 

452488 

7-12 

981812 

.62 

470676 

7-74 

529324 

32 

291' 28374  93890 

452910 

7.11 

981774 

.62 

471141 

7-73 

523839 

31 

30 

28402  95S82 

453342 

7.10 

981737 

.62 

471605 

7-73 

528395 

30 

SI 

28420  o5^-4 

9-453768 

7.10 

9.981699 

•63  9.472068 

7.72 

10-527932 

29 

32.  2S457  9^^^-'^ 

454194 

7.09 

981662 

•63 

472332 

7'7i 

527468 

23 

33  i  28435  95357 

454619 

7.08 

981620 

.63 

472993 

7.71 

527005 

27 

34i:2S5i3  95849 

455044 

7-07 

981587 

•  63 

473437 

7.70 

526543 

26 

351  2S54I  93841 

450469 

7.07 

981049 

•  63 

473919 

7.69 

526081 

25 

86!  28569  9^332 

455S93 

7.06 

981512 

•63 

474381 

7.69 

525619 

24 

37  i  28597  93824 

4563i6 

7.00 

931474 

.63 

474842 

7.68 

525i58 

23 

38  1  28625  95816 

39  1  28652  95807 

406739 

7-04 

981436 

•63 

475303 

7.67 

524697 

22 

407162 

7-04 

981399 

•  63 

473763 

7.67 

524237 

21 

40 

28680  95-99 

457584 

7 -03 

981361 

.63 

476223 

7.66 

523777 

20 
19 

41 

28708  95-91 

9.453006 

7.02 

9-981323 

•63  ,'9.476683 

7.60 

10-523317 

42  28-36  95-82 

453427 

7.01 

981280 

.63 

477142 

7.60 

522858 

13 

43'  28764  95774 

453348 

7.01 

931247 

•  63 

477601 

7-64 

522399 

17 

44  2S792  95766 

459268 

7.00 

981209 

.63 

478059 

7-63 

521941 

16 

451  28820  95757 

45o653 

6-99 

981171 

•63 

478517 

7-63 

521483 

15 

46 

28847  93^49 

460108 

6-98 

981133 

•  64 

478975 

7.62 

521025 

14 

47 

,28875  95-40 

460527 

6.98 

981095 

.64 

479432 

7.61 

520563 ] 13 

48 

28903  95732 

460946 

6-97 

981057 

.64 

479889 

7-6i 

5201 1 1  12 

49 

28931  95724 

461 364 

6-96 

981019 

•64 

480345 

7-60 

519655  111 

50 

28959  95^i5 

461782 

6-95 

980981 

.64 

480801 

7-59 

519199  10 

51  j  28987  93707 

9.462199 

6.95 

9-980942 

•64 

:9-48i257 

]'U 

10.518743 

9 

52  1  29010  95698 

462616 

6-94 

980904 

.64 

481712 

518288 

8 

53  '   29042  q56qo 

463o32 

6-93 

980^66 

.64 

482167 

^1^ 

517833 

7 

54 

29070  95681 

463448 

6-93 

980827 

.64 

482621 

7-57 

517379 

6 

55 

29098  95673 

463864 

6-92 

980789 

.64 

483075 

7-56 

516920 

5 

56 

,29126  95664 

464279 

6-91 

980750 

•64 

483529 

7-30 

516471 

4 

57 

129154  95'j56 

464694 

6-90 

9807  1 2 

•64 

4839S2 

7-55 

5 1 601 8 

3 

58 

29182  95647 

465 1 08 

6.Q0 

980673 

.64 

484435 

7-54 

5i5565 

2 

59 

29209  9563g 

465522 

980635 

.64 

484887 

7-53 

5i5ii3 

1 

60 

'29237  95530 

463935 

6-88 

980596 

.64 

485339 

7.53 

5i466i 

0 

N.  COS.  X.  sine. 

L.  COS. 

D.l" 

j  L.sine.  j 

KcoL 

1  D.l" 

L.  tang. 

73^                         1 

TRIGONOMETRICAL    FUNCTIONS. — 17°. 


47 


Nat.  Functions. 

Logarithmic  Functions  +  10.             1 

_1 

N.sine.  N.  cos. 

L.  sine. 

D.1" 

L.  COS.  JD.l" 

L.  tang. 

D.  1" 

L.  cot 

1 

0 

29237 

95630 

9.460935 

6-88 

9-980596  -64 

9-485339 

7-55 

io-5i466i!60 

1 

29265 

95622 

466348 

6-88 

980508 1 

.64 

485791 

7.5a 

514209 [59 

2 

29293  j  90613 

466761 

6-87 

980519 

■  65 

486242 

7.5, 

5i3758;5S 

3 

29321  956o5 

467173 

6-86 

980480 

•60 

486693 

7-5i 

513307; 57 

4 

29348  95596 

467585 

6-85 

980442 

•65 

487143 

7-5o 

512857! 56 

5 

29376  95588 

467996 

6-85 

980403 

-65 

487593 

7-49 

512407 

55 

6 

29404  95579 

468407 

6-84 

98o364 

•  65 

488043 

7-49 

5,1957 

54 

7 

29432  95571 

468817 

6-83 

980325 

•  65 

4S8492 

7-48 

5iioo8 

53 

8 

29460  95562 

469227 

6-83 

980286 

•  65 

488941 

7-47 

5iio59J52 

9 

29487  95554 

469637 

6-82 

980247 

.65 

489390 

7-47 

5io6,u  51 

10 
11 

295i5|  95545 

470046 

6-8i 

980208 

.65 

•  489838 

.7.46 

510162  150 

29543 

95536 

9-470455 

6-80 

9-980169 

.65 

9.490286 

7-46 

10-5097,4 

49 

12 

29571 

95528 

470863 

6-80 

980130 

•  65 

490733 

7-45 

509267 

43 

13 

29399 

95519 

471271 

'6-79 

980091 

•  65 

491180 

7-44 

308820 

47 

14 

29626 

955,1 

471679 

6-78 

980052 

.65 

491627 
492073 

7-44 

508373 

46 

15 

29604 

905o2 

472086 

6-78 

980012 

•  65! 

7-43 

507927 

45 

16 

29682 

95493 

472492 

6-77 

979973 

.65 

492519 

7-43 

50748, 

44 

17 

29710 

95485 

472898 

6-76 

979934 

.66' 

492960 

7-42 

507035 

43 

18 

29737 

95476 

473304 

6-76 

979895 

.66 

493410 

7-41 

506590 

42 

19 

29765 

95467 

473710 

6-75 

979855 

.66 

493854 

7-40 

506146 

41 

20 
21 

29793  1  95459 

4741 1 5 

6-74 

979816 

.66' 

. 494299 

7-40 

5o57o, 

40 

29821 

95450 

9-474519 

6.74 

9-979776 

.66 

9-494743 

7-40 

,o-5o5257 

89 

22 

29849 

95441 

474923 

6-73 

979737 

.66 

495186 

7-39 

504814 

83 

23 

29876 

95433 

475327 

6.72 

979697 

.66 

495630 

7-38 

504370 

37 

24 

29904]  95424 

473730 

6-72 

919658 

.66' 

496073 

7-37 

503927 

86 

25 

29932  I  95415 

476i33 

6-71 

979618 

•  661 

49651 5 

7-37 

5o3485 

35 

26 

29960;  95407 

476536 

6-70 

979079 

.66 

496957 

7-36 

5o3o43 

34 

27 

29987  95398 

476938 

6-69 

979539 

.66 

497399 

7-36 

5o26oi 

33 

28 

3ooi5!  95389 

477340 

6-69 

979499 

.66 

497841 

7-35 

5o2i59 

32 

29 

30043 i  95380 

477741 

6-68 

979409 

.66 

498282 

7-34 

501718 

31 

30 
81 

30071 
30098^ 

95372 

478142 

6-67 

979420 

.66 

498722 
9.499163 

7-34 
7-33 

50,278 
io-5oo837 

80 
2y 

95363 

9-478542 

6-67 

9-979380 

.66 

32 

30126  I  95354 

478942 

6-66 

979340 

.66 

499603 

7-33 

5oo397 

28 

33 

3oi54i  95345 

479342 

6-65 

979300 

.67 

500042 

7.32 

499908 

27 

34 

30182;  95337 

479741 

6-65 

979260 

•67 

500481 

7-3i 

499319 

26 

35 

30209;  95328 

480140 

6-64 

•  979220 

-67 

000920 

7-3. 

499080 

25 

36 

30237  \   95319 

480539 

6-63 

979180 

•67' 

5oi359 

7.30 

498641 

24 

87 

3o265;  95310 

480937 

6-63 

979140 

•67 

501797 

7.30 

498203 

.23 

38 

30292 1  95301 

481334 

6-62 

979100 

.67 

502235 

7.29 

497765 ! 22  1 

89 

3o32oi  95293 

481731 

6-6i 

979059 

.67 

502672 

7.28 

497328 

21 

40 

3o348|  95284 

482128 

6.61 

979019 

•67 

5o3i09 

7.28 

496891 
10-496454 

20 

^^1 

30376 i  90270 

9-482525 

6-60 

9-978979 

.67; 

9-5o3546 

7-27 

42 

3o4o3i  95266 

482921 

6.59 

978939 

.67 

003982 

7-27 

496018 

18 

431 

3o43i  i  90257 

4833 16 

6.59 

978898 

-67 

5o44 1 8 

7-26 

495582 

]7 

44 1 

30459!  90248 

483712 

6-58 

978858 

.67' 

504854 

7-25 

490146 

16 

45! 

30486;  95240 

484107 

6.57 

978817 

•671 

505289 

7-25 

4947" 

15 

46  1 

3o5i4 

9523i 

484501 

6-57 

978777 

•67: 

505724 

7-24 

494276 

14 

47 

30042 

95222 

4S4895 
485289 

6-56 

978736 

.67  1 

5061O9 

7-24 

493841 

13 

48 

30570 

95213 

6-55 

978696 

•68, 

506593 

7-23 

493407 

12 

49 

3o597 

95204 

485682 

6-55 

978655 

-68 

507027 

7-22 

492973 

11 

«0 

3o625 

95195 

486070 

6-54 

978615 

•68! 

507460 

7-22_ 

492040 
10-492107 

10 
9 

3o653}  95186 1 

9-486467 

6-53 

9-978574 

•  68 

9-507893 

7-21 

52 

3o68o|  95177 

486860 

6-53 

978533 

.68, 

5o8326 

7-21 

491674 

8 

53 

30708!  95168 

487201 

6-52 

978493 

-68* 

508759 

7-20 

491241 

7 

54 

30736 

90159 

487643 

6.5i 

978402 

-68  i 

509191 

7-19 

49080Q 
490378 

6 

55 

30763 

95i5o 

488034 

6-5i 

978411 

-68! 

509622 

?:;? 

5 

56 

30791 

95142 

488424 

6-5o 

978370 

-68 

5 10004 

489946 

4 

57 

30819 

95i33 

488814 

6-5o 

978329 

-68! 

5 1 0485 

7-,8 

489015 

3 

58 

30846 

95124 

489204 

6-49 
6-48 

978288 

-68 

5 ! 09 1 6 

7-'7 

489084 

2 

59  1 

30874 

90115 

489093 

978247 

-68 

5ii346 

7.16 

488604 

1 

601 

30902 

95106 

489982 

6-48 

978206 

.68 

511776 

7.16 

488824 

0 

N.  COS.  iN.  sine.] 

L.  COS. 

D.  1" 

L.  sine. 

L.  cot. 

D.  1" 

L.  tang. 

' 

72^                                                              1 

48 


TRIGONOMETRICAL  FUNCTIOl^S. — 18" 


Nat.  Functions. 

Logarithmic  Functions  +  10.             1 

' 

•N.sineJ  N.  cos. 

L.  sine. 

D.  1' 

L.  COS. 

D.l" 

L.  tang. 

D.l" 

L.  cot 

0 

30902  93106 

9.489982 

6.48 

9-978206 

•  68 

9.511776 

7.16 

10.488224 

60 

1 

30929 1 90097 

490311 

6 

48 

978165 

.68 

5l2206 

7.16 

487794 

59 

2 

30907  I  90088 

490759 

6 

47 

^''I'l^ 

.68 

512635 

7-15 

487365 

58 

S 

30980 1 90079 

491147 

6 

46 

978083 

.69 

5i3o64 

7-14 

486936 

57 

4 

31012190070 

491535 

6 

46 

978042 

.69 

513493 

7-U 

486507 

5(5 

6 

3 1 040 

95061 

491922 

6 

45 

978001 

.69 

513921 

7-13 

486079 

55 

6 

3 1 068 

95o52 

492308 

6 

44 

977959 

.69 

514349 

7. .3 

485651 

54 

7 

31095 

95043 

492695 

6 

44 

977918 

.69 

514777 

7.12 

485223 

53 

8 

31123 

90033 

493081 

6 

43 

911^11 

.69 

5i02o4 

7.12 

484796 

52 

9 

3II0I 

95024 

493466 

6 

42 

977835 

.69 

5i563i 

7.11 

484369 

51 

10 
11 

31178 

95oi5 

493851 

6 

42 

977794 

.69 

5i6o57 

7.10 

483943 

50 
49 

3 1 206 

95006 

9.494236 

6 

41 

9.977702 

.69 

9.516484 

7.10 

io.4835i6 

12 

3.233 

94097 

494621 

6 

41 

977711 

.69 

516910 

7-09 

483090 

48 

13 

3 1 26 1  949><8 

495oo5 

6 

40 

977669 

.69 

517335 

U 

482665 

47 

U 

31289,94979 

495388 

6 

39 

977628 

.69 

517761 

482239 

46 

15 

3i3i6 

94970 

495772 

6 

39 

977586 

.69 

518180 

7.08 

481815 

45 

16 

3i344 

94961 

496154 

6 

38 

977544 

.70 

518610 

7-07 

481390 

44 

17 

3i372 

94952 

496037 

6 

37 

9775o3 

.70 

519034 

7-o6 

480966 

43 

18 

3i399 

94943 

496919 

6 

37 

977461 

.70 

519458 

7.06 

480042 

42 

19 

31427 

94933 

497301 

6 

36 

977419 

.70 

519882 

7.05 

480118 

41 

20 

31404 

94924 

497682 

6 

36 

977377 

.70 
•70 

52o3o5 

7.05 

^47969! 

40 
39 

21 

31482 

94915 

9.498064 

~b 

35 

9.977335 

9.520728 

7-04 

10.479272 

"22 

3i5io 

94906 

498444 

6 

34 

977293 

.70 

521101 

7-03 

478849 

38 

23 

31037 

94^97 

498825 

6 

34 

977201 

•70 

521573 

7.03 

478427 

37 

24 

31065 

94888 

499204 

6 

33 

977209 

•70 

521995 

7.03 

478005 

36 

.25 

31D93 

94878 

499584 

6 

32 

977167 

.70 

522417 
522838 

7.02 

477583 

35 

26 

3i62o 

94869 

499963 

6 

32 

977125 

.70 

7.02 

477«62 

34 

27 

31648 

94860 

5oo342 

6 

3i 

977083 

.70 

523259 

7.01 

476741 

S3 

28 

31670 

9485 1 

500721 

6 

3i 

977041 

.70 

523680- 

7.01 

476320 

32 

29 

3 1 703 

94842 

501099 

6 

3o 

976999 

.70 

524100 

7.00 

475900 

31 

30 

31730 

94832 

501476 

6 

29 

976907 

.70 

524520 

6.99 

475480 

SO 

31 

31708 

94823 

9.001804 

6 

29 

9.976914 

•70  1 

9.524939 

6.99 

10-475061 

29 

32 

31786 

94814 

50223l 

6 

28 

976872 

•71 

525359 

6.98 

474641 

28 

33 

3i8i3 

94805 

502607 

6 

28 

976830 

•71 

525778 

6.98 

474222  27 

34. 

31841 

94795 
94786 

502984 

6 

27 

976787 

•71 

526197 

6.97 

4738o3  26 

35 

3 1 868 

5o336o 

6 

26 

976745 

•71 

5266 1 5 

6-97 

473385 

25 

36 

31896 

94768 

■  5o3730 

6 

26 

976702 

•71 

527033 

6.96 

472967 

24 

37 

31923 

5o4tio 

6 

25 

976660 

•71 

527451 

6.96 

472049 

23 

38 

31951 

94758 

504485 

6 

25 

976617 

•71 

527868 

6.95 

472132 

22 

39 

31979 

94749 

504860 

6 

24 

976574 

•71 

528285 

6.95 

471713 

21 

40 
41 

32006 

94740 

5o5234 

6 

23 

976532 

•71 

528702 

6.94 

471298 

20 

32034 

94730 

9.5o56o8 

"T 

23 

9.976489 

•7» 

9.529119 

6.93 

10-470881 

19" 

42 

32061 

94721 

005981 

6 

22 

976446 

•71 

529530 

6.93 

470465  18 

43 

32089 

94712 

5o6354 

6 

22 

976404 

.71 

529950 

6.93 

470o5o  j 17 

44 

321 16  j  94702 

^  506727 

6 

21 

976361 

•71 

53o366 

6.92 

469634  16 

45 

32144  94693 

507099 

6 

20 

976318 

•71 

530781 

6.91 

469219 
468804 

15 

46 

32171  1 94684 

507471 

6 

20 

976275 

•71 

531196 

6-91 

14 

47 

32199  1  94674 

507843 

6 

«9 

976232 

•72 

531611 

6-90 

468380 
467975 
467561 

13 

48 

32227  94665 

508214 

6 

»9 

976189 

.72 

532025 

6.90 
6.89 

12 

49 

32204  94656 

5o8585 

6 

18 

976146 

.72 

532430 
532853 

11 

50 

32282  94646 

508956 

6 

18 

976103 

.72 

6.89 

467147 

10 

51 

32309:94637 

9.509326 

6 

n 

9.976060 

•72 

9.533266 

6.88 

10.466734 

9 

52 

32337  94627 

509696 

6 

16 

976017 

.72 

533679 

6.88 

466321 

8 

53 

32364  94618 

5 10065 

6 

16 

975974 

.72 

534092 

5.87 

465908 

7 

54 

32392  04609 

510434 

6 

i5 

975930 

.72 

534504 

•;  6.87 

465496 
465o84 

6 

55^32419^94599 

5io8o3 

6 

i5 

975887 

.72 

534916 

'6.86 

5 

56  I:  32447  I  q45qo 

511172 

6 

14 

973844 

.72 

535328 

•  9.86 

464672 

4 

57 

32474  94080 

5ii54o 

6 

i3 

975800 

.72 

535739 

6.85 

464261 

3 

58 

32002  94071 

511907 
512275 

6 

i3 

970757 

.72 

536 1 5o 

6-85 

463850 

2 

59 

32529  94561 

6 

12 

970714 

.72 

536561 

6.84 

463439 
463028 

1 

60 

32557 j 94552 

512642 

6-12 

975670  • 

.72 

1  536972 

6-84 

0 

|N.  COS.  N.  sine. 

L.  COS. 

D.l" 

L.  sine. 

1  L.  cot 

D.l' 

L.  tang. 

1 

•71° 

1 

TRIGONOMETRICAL   F  TXCTIOXS. — 19". 


49 


Nat.  Functions. 

Logarithmic  Functions  -t-  10.              1 

'   1 

1 

N.sine.!N.  cos.| 

L.  sine. 

D.  1"   L.  COS.  |] 

D.l" 

;  L.  tang. 

D 1."    L.  cot.     1 

o! 
1' 

2; 

3 

4 

\ 

7 

8 

9 

10 

11 
12 
13 
14 
15 
16 
17 
18 

io 

20 

"2r 

23 
23 
24 
25 
26 
27 
23 
29 
30 

32557 1 

32584 

32612 

33639 

32667 

32694 

32722 

32749 
32777 
32804 
32832 

32859 
32887 
32914 
32942 
32969 
132997 
33024 
:33o5i 
1 33079 
33io6 

94552 
94542 
94533 
94523 
94514 
94504 
94495 
94485 
94476 
94466 
94457 

9-512642 
5 1 3009 
5x3375 
5i374i 
514107 
514472 
514837 

5l5202 

5 1 5566 
5i593o 
516294 

6-12 

6.11 
6.11 
6.10 
6-09 
6.09 
6-o8 
6-08 
6-07 
6.07 
6-06 

9.973670 

973539 
973496 
975432 
973408 
975365 
975321 

975277 
975233 

•73 
•73 

1 

9-536972 
537382 
537791 
538201 
5386x1 
539020 
539429 
539837 
540245 
540653 
541061 

6-84 
6-83 
6.83 
6-82 
6-82 
6-81 
6-8i 
6-80 
6-80 
6-79 
6-79 

10-463028  60 
462618  59 
402208  1  58 
461798  1  57 
461389156 
460980 1 55 
460371  1 54 
460163  153 
439.755  j  52 
459347  1  51 
435939  50 

94447 
94438 
94428 
94418 
94409 
94399 
94390 
94380 
94370 
94361 

9.516657 
517020 
517382 
517743 
518107 
518468 
518829 
519190 
519551 
519911 

6-o5 
6-05 
6.04 
6.04 
6.o3 

6-02 

6.02 
6-01 
6-01 
6.00 

9-975189 
973143 
975101 
97^037 
973013 
974969 
974923 
974880 
974836 
974792 

1 

•74 
•74 
•74 
•74 
•74 
•74 
•74 
•74 
•74 
•74 
•74 
•74 
•74 
•74 
•75 

9.541468 
541875 
542281 
542688 
543094 
543499 
543905 
544310 
544715 
545119 

6.78 
6-78 
6.77 
6.77 
6.76 
6.76 
6.75 
6.73 
6-74 
6-74 

10-438532 
458123 

437719 
437312 
436906 
456301 
456095 
433690 
455285 
43488 X 

49 
48 
47 
46 
45 
44 
43 
42 
41 
40 

33i34 
33i6i 
'83189 
j 33216 
33244 
1 33271 
1 33298 

: 33326 

! 33353 

|3338i 

94351 
94342 
94332 
94322 
943 1 3 
943o3 
94293 
94284 
94274 
94264 

9.520271 
52063 1 
520990 
521349 
521707 
522066 
522424 
522781 
523i38 
523495 

6-00 
5-99 

5.98 
5.?8 
5-97 
5.96 
5.96 
5.93 
5-93 

9.974748 
974703 
974659 
974614 
974570 
974523 
974481 
974436 
974391 
974347 

9-545524 

545928 
546331 
546735 
547138 
547540 
547943 
548345 
548747 
549149 

6-73 
6.73 
6.72 
6.72 
6.71 
6.71 
6-70 
6-70 
6-69 
6-69 

10.454476 

434072 
453669 
453265 
432862 

432400 
452037 

431655 

431233 

43o85i 

39 
33 
37 
36 
35 
34 
33 
82 
31 
30 

29 
28 
27 
26 
25 
24 

r3 

22 
21 
20 

31 
32 
S3 
34 
35 
86 
37 
38 
39 
40 

'  33408" 
1 33436 
' 33463 
'  33490 
i335i8 
33545 
1 33573 
'■■  33600 
133627 
1 33655 

94254 
94245 
94233 
94225 
94215 
94206 
94196 
94186 
94176 
94167 

9.523«52 
524208 
524564 
524920 
525275 
525630 
525984 
526339 
526693 
527046 

5-94 
5.94 
5.93 
5.93 
5.92 
5.91 
3.91 
5.90 
5-90 
5-89 

9-974302 
974257 
974212 
974167 
974122 

974077 
974032 
973987 
973i)42 
973897 

•75 
•75 
•75 
•75 

3 
I 

9-549530 
549951 
55o352 
550752 
55ii52 
55x552 
551952 
552351 
552750 
553x49 

6-68 
6-68 
6-67 
6-67 
6-66 
6-66 
6-65 
6-65 
6-65 
6-64 

10-430450 

430049 
449648 
449248 

445843 

44S448 

448048 
447649 
447250 

440&3I 

41 
42 

43 
44 
45 
46 

47 
43 
49 
50 

i  33682 
133710 
133737 

; 33764 
33792 
33819 

33846 

33874 
|!  33901 
433929 

94137 
94147 
94137 
94127 
94118 
94108 
94098 
94088 
94078 
94068 

9.527400 
527753 
528io5 
528458 
528810 
529161 
529513 
529864 
53021 5 
53o565 

5-89 
5-88 

^  5.88 
5-87 
5.87 
5.86 
5.86 
5.85 
5.85 
5-84 

9-973852 
973807 
973761 
973716 
973671 
973623 
973530 
973535 
973489 
973444 

•75 

.76 
.76 
•76 
.76 
-76 
-76 
-76 

9-553548 
553946 
554344 
554741 
555x39 
555536 
555933 
556329 
556725 

1  55712X 

6.64 
6.63 
6.63 
6-62 
6.62 
6-6x 
6-6x 
6-60 
6.60 
6-59 

10-446452 
446054 
443036 
443239 
444861 
444i64 
444067 
443071 
443275 
442579 

19 
18 
17 
16 
15 

i^ 

12 
11 
10 

51|!339d6 

52  i|  33983 

53  |340M 

54  34038 

55  !i  34065 
50  ; 34093 

57,;  34120 

58  I  34147 
591:34175 

60  ; 34202 

94o58 
94049 
94039 
94029 
94019 
94009 
93999 
93989 
93979 
93969 

9-53o9i5 
531265 
53i6i4 
531963 
532312 
532661 
533009 
533357 
533704 
534052 

5-84 
5-83 
5.82 
5-82 
5-81 
5.81 
5-80 
5-80 
5-79 
5-78 

9.973398 
973332 
973307 
973261 
973215 
973169 
973124 
973078 
973o32 
973986 

-76 
.76 
-76 
-76 
•76 
-76 
-76 
-76 

•77 
•77 

9-557517 
557913 
5583o8 
55S702 
559097 
539491 
559885 

56 1 066 

6.59 
6.59 
6-58 
6-58 
6.57 
6.57 
6.56 
6-56 
6-55 
6.55 

10-442483 
442087 
44I6Q2 
441298 
440903 
440309 
4401x5 
439721 
439327 
435934 

9 
8 
7 
6 
5 
4 
3 
2 
1 
0 

N.  COS. 

N.6ine.|  L.  COS. 

D.l" 

L.  sine. 

:  L.  cot. 

D.l" 

L.  tang.  1  '  1 

70°                         1 

50 


TRIGONOMETRICAL   FUXCTIOXS. — 20° 


Nat.  Fvnctions. 

LooARiTUMic  Functions  +  10.             1 

T 

3 

4 

5 

6 

7 

S 

9 
10 
11 
12 
13 
14 
15 
Id 
17 
13 
19 
20 

N.sme.|N.cos. 

L.sine.  'd 

1"  j  L.  COS. 

D.l" 

L.  tang. 

D.l" 

L.  cot 

34202 

34229 
34207 
34284 
343 1 1 
34339 
34366 
34393 
34421 
34448 
34475 

93969 
93909 
93949 
93939 
93929 
93919 
93909 
93899 
93889 

93879 
93869 

9.534052 

534399 
534745 
530092 
535438 
535783 
536129 
536474 
5368i8 
537163 
537507 

5 
5 
5 
5 
5 
5 
5 
5 
5 
5 
5 

78 
77 
77 
77 
76 
76 
75 
74 

II 

9.972986 
972940 
972894 
972848 
972802 
972755 
972709 
972663 
972617 
972570 
972524 

•77 
•77 
•77 
•77 
•77 
•77 
•77 
•77 
•77 
•77 
•77 

9-56io66 
561409 
66i85i 
562244 
562636 
563028 
563419 
563811 
564202 
564592 
564983 

6.55 
6.54 
6.54 
6.53 
6.53 
6.53 
6.52 
6-52 
6-51 
6.5i 
6.5o 

10.438934 
438541 
433149 
437756 
437364 

436189 
435798 
435408 
435017 

60 
59 
53 
57 
56 
55 
54 
53 
52 
51 
50 

345o3 

' 34530 
34507 

1 34084 
34612 
34639 

' 34666 
34694 
34721 

,34748 

93809 
93849 
93839 
93329 
93319 
93809 

93799 
93789 

93779 
93769 

9-537851 
538194 
538538 
538880 

•  539223 
539565 
539907 
540249 
540590 
540931 

5 
5 
5 
5 
5 
5 
5 
5 
5 
5 

72 
72 
71 
71 
70 
70 

68 

9-972478 
972431 
972385 
972338 
972291 
972245 
972198 
972i5i 
972105 
972058 

.78 
-78 

9.560373 
565763 
5661 53 
566542 
566932 
567320 

563486 
568873 

6.5o 
6.49 
6.49 
6-49 
6-48 
6.48 
6-47 
6-47 
6.46 
6-46 

10-434627 
434237 
433847 
433458 
433068 
432680 
432291 
431902 
431014 
431127 

49 
4S 
47 
46 
45 
44 
43 
42 
41 
40 

•2i 
22 
23 
24 
25 
26 
27 
25 
29 
30 
ol 
32 
33 
34 
35 
36 
37 
33 
39 
40 

34775 
■  34803 
34830 
34857 
34884 
3491^ 
34939 
34966 
34993 
35o2i 

93759 
9374S 
9373b 

93708 
93698 
93688 
93677 
93667 

9-541272 
541613 
541953 
542293 
542632 
542971 
543310 
543649 
543987 
544325 

5 
5 
5 
5 
5 
5 
5 
5 
5 
5 

66 
65 
65 
64 
64 
63 
63 

9-972011 
971964 
971917 
971870 
971823 
971776 
971729 
9716S2 
971635 
971588 

t 
:?^ 

•79 
•79 
•79 
•79 

9.569261 
069648 
570035 
570422 
570809 
571,95 
571581 
571967 
572352 
572733 

6.40 
6.40 
6.45 
6-44 
6-44 
6.43 
6.43 
6.42 
6.42 
6.42 

10-430739 
43o3o2 
429965 
429578 
429191 
428800 

428419 
428033 
427648 
427262 

39 
38 
37 
36 
35 
34 
33 
32 
31 
30 

35048 
35075 
35102 
,35i3o 
3oi57 
35i84 
352II 
35239 
35266 
35293 

93657 
93647 
93637 
93626 
93616 
93606 
93596 
93585 
93070 
93565 

9-544663 
545000 
545338 
545674 
54601 1 
546347 
546683 
547019 
547354 
547689 

5 
5 
5 
5 
5 
5 
5 
5 
5 
5 

62 
62 
6i 
61 
60 
60 

% 

58 

58 

9.971540 
971493 
971446 
971398 
97i3oi 
97i3o3 
971256 
971208 
971161 
971113 

•79 
•79 
•79 
•79 
•79 
•79 
•79 
•79 
•79 
•79 

9.573123 
573507 
573892 
574276 
574660 
575044 
575427 
570810 
576193 
576576 

6.41 
6.41 
6.40 
6-40 
6.39 
6.39 
6-39 
6-33 
6-38 
6.37 

10-426877 
426493 
426108 
425724 
425340 
424956 
424373 
424190 
423807 
423424 

29 
28 
27 
26 
25 
24 
23 
22 
21 
20 

41 
42 
43 
44 
45 
46 
47 
4S 
49 
50 

; 35320 
1 35347 
; 35375 
! 35402 
35429 
35456 
35484 
;355ii 
135533 
135565 

93555 
93544 
93534 
93524 
93514 
93oo3 
93493 
93483 
93472 
93462 

9.548024 
548359 
548693 
549027 
549360 
549693 
550026 
55o309 
550692 
55io24 

5 
5 
5 
5 
5 
5 
5 
5 
5 
5 

56 
56 
55 
55 
54 
54 
53 
53 

9-971066 
971018 
970970 
970922 
970874 
970827 
970779 
970731 
970683 
970635 

.80 
.80 
.80 
-80 
•80 
.80 
.80 
.80 
•80 
-80 

9.576908 
577341 
577723 
578104 
578486 
578867 
579248 

%tll 
58o389 

6.37 
6.36 
6-36 
6.36 
6.35 
6.35 
6.34 
6.34 
6/34 
6.33 

10.423041 
422659 
422277 
421896 
42i5i4 
421133 
420752 
420371 

419991 
419611 

19 
IS 
17 
16 
15 
14 
13 
12 
11 
10 

51 
52 
53 
54 
55 
56 
57 
58 
59 
60 

,35592 
! 35619 
i  35647 
135674 
j 35701 
35728 
35755 
30782 
35810 
35837 

93452 
93441 
93431 
93420 
93410 
93400 
93389 

93858 

9.501306 
551687 
552018 
552349 
552680 
553010 
555341 
553670 
554000 
554329 

5 
5 
5 
5 
5 
5 
5 
5 
5 
5 

52 
52 
52 

5i 
5i 

5o 
5o 

49 
49 
48 

9-970586 
970538 
970490 
970442 
970394 
970345 
970297 

970249 
970200 
970152 

.80 
.80 
.80 
.80 
.80 
.81 
-81 
.81 
.81 
.81 

9.580769 
581149 
58i528 

582286 
582665 
533043 
583422 
583300 
584177 

6.33 
6-32 
6-32 
6-32 
6-31 
6-3i 
6-30 
6-3o 
6-29 
6-29 

10.419231 
4i83oi 
418472 
418093 
417714 
417335 
416957 
416573 
416200 
41 5823 

9 
8 
7 
6 
5 
4 
8 
2 
1 
0 

t 

N.  COS. 

X.sin*. 

L.  COS. 

D.  1" 

L.8ine.  i    1 

L.  cot. 

D.l" 

L.  tang. 

69                          1 

TRIGONOMETRICAL    FUNCTIONS.— 21< 


51 


Nat.  Functions. 

Logarithmic  Functions  +  10. 

'  |!N.8lne.|N.co8 

L.  sine. 

D.  1" 

L.  COS. 

D.l" 

1  I*  tang. 

D.l" 

L.  cot.  ! 

0 

35837  j  93358 

9-554329 

5.48 

9.970152 

.81 

9-584177 

6-29 

10 ^41 5823  1  60 

1 

35864 ! Q3348 

554658 

5-48 

970103 

-81 

584555 

6-29 

410445  :  5y 

2 

35891 

93337 

554987 

5-47 

970055 

-81 

584932 

6-28 

4i5o68  '   58 

3 

35918 

93327 

55531 5 

5-47 

970006 

.81 

585309 

6-28 

414691 

57 

4 

35945 

93316 

555643 

5-46 

969957 

•81 

585686 

6-27 

4i43i4 

56 

5 

35973 

93306 

555971 

5.46 

969909 

•81 

586062 

6^27 

413938 

55 

6 

36000 

93295 

556299 

5.45 

969860 

.81 

586439 

6.27 

4i356i 

54 

7 

36027 

93285 

556626 

5-45 

9698 1 1 

.81 

58681 5 

6^26 

4i3i85 

53 

8 

36o54 

93274 

5^6953 

5-44 

969762 

-81 

581190 

i  6^26 

412810 

52 

9 

36o8i 

93264 

557280 

5-44 

Q69714 

.81 

587566 

6^25 

412434 

51 

10 

36 1 08 

93253 

557606 

5.43 

969665 

•81 

587941 

6-25 

412059 

50 

11 

36i35 

93243 

0-557932 

5-43 

9.969616 

-82 

9-588316 

6^25 

10-411684  ;  4j 

12 

36162 

93232 

558258 

5-43 

969567 

.82 

588691 

6-24 

4ii3o9  1  48 

13 

36190 

93222 

558583 

5-42 

969018 

-82 

589066 

6^24 

410934 

47 

14 

36217 

93211 

558909 

5-42 

969469 

.82 

589440 

6^23 

4io56o 

46 

15 

36244 

93201 

559234 

5.4. 

969420 

-82 

589814 

6-23 

410186 

45 

16 

36271 

93190 

559558 

5-41 

969370 

.82 

590188 

6^23 

409812 

44 

17 

36298 

93180 

559883 

5-40 

969321 

.82 

590562 

6^22 

409438 

43 

18 

36325 

93169 

560207 

5.40 

969272 

.82 

590935 

6^22 

409065 

42 

19 

36352 

93i5g 

56o53i 

5.39 

969223 

-82 

591308 

6-22 

408692 

41 

20 

36379 

93148 

56o855 

5.39 

969.73 

.82 

591681 

6^21 

408319 

4<i 

21 

36406 

93137 

9-561178 

"5.38 

9-969124 

-82 

9.592054 

6-21 

10-407946 

Sy 

22 

36434 

93127 

56i5oi 

5-38 

969073 

.82 

592426 

6^20 

407574 

38 

23 

36461 

93116 

561824 

5-37 

969025 

.82 

592798 

6^20 

407202 

37 

24 

36488 

93106 

562146 

5-37 

968976 

.82 

593171 

6^19 

406829 

36 

25 

365i5 

93095 

562468 

5-36 

968926 
968877 

.83 

593542 

6.19 

406408 

35 

2(5 

j  36542 

93084 

562790 

5-36 

.83 

593914 

6.18 

406086 

34 

27 

36569 

93074 

563112 

5-36 

968827 

.83 

594285 

6.18 

40571 5 

33 

28 

36596 

93o63 

563433 

5-35 

968777 

.83 

594656 

6.i8 

405J44 

32 

29 

36623 

93o52 

563755 

5.35 

968728 

.83 

595027 

6^17 

404973 

31 

30 

36650 

93042 

564075 

5.34 

968678 

•83 

595398 

6^17 

404602 

30 

2y 

31 

36677 

93o3i 

9.564396 

5.34 

9.968628 

-83 

9^595768 

6.17 

10^404232 

32 

36704 

93020 

564716 

5.33 

968578 

-83 

596138 

6.i6 

4o3862 

28 

33 

36731 

93010 

565o36 

5-33 

968528 

-83 

596508 

6^i6- 

403492 

27 

34 

36758 

92999 

565356 

5-32 

968479 

.83 

596878 

6.16 

4o3i22 

26 

35 

36785 

929S8 

565676 

5.32 

968429 

-83 

■  597247 

6.i5 

402753 

25 

3(i 

; 36812 

92978 

565995 

5-31 

96S379 

•83 

597616 

6-15 

402384 

24 

37 

j  36839 

92967 

566314 

5-3i 

968329 
968278 

-83 

597985 

6-15 

40201 5 

23 

83 

136867 

92956 

566632 

5.31 

•83 

598354 

6^14 

401646 

22 

39 

36894 

92945 

566951 

5.30 

968228 

-84 

598722 

6^14 

401278 

21 

40 

36921 

92935 

567269 

5-3o 

968178 

-84 

599091 

6-13 

400909 

20 

41 

36948 

92924 

9.567587 

5.29 

9-968128 

•  84 

9.599459 

6-i3 

10-400541 

19 

42 

36975 

92913 

567904 

5-29 

968078 

.84 

599827 

6-i3 

400173 

18 

43 

37002 

92002 
92892 

568222 

5.28 

968027 

.84 

600 1 94 

6^12 

399S06 

17 

44 

37029 

568539 

5.28 

9^7977 

.84 

6oo562 

6-12 

399438 

16 

45 

37056 

92881 

568856 

5-28 

967927 

.84 

600929 

6^11 

399071 

15 

46 

37083 

92870 

569172 

5.27 

967876 

.84 

601296 

611 

39.^704 

14 

47 

37110 

92859 

569488 

5.27 

967S26 

.84 

601662 

6^11 

398338 

13 

48 

37137 

92849 

569804 

5-26 

967775 

.84 

602029 

6-10 

39^971 

12 

49 

37164 

92838 

570120 

5-26 

967725 

.84 

602395 

6-10 

397605 

11 

50 

37191  92S27 

570435 

5-25 

967674 

.84 

602761 

6^io 

397239 

10 

51 

37218  92816 

9-570751 

5.25 

9.967624 

.84 

9-603127 

6-09 

10-396873 

7 

52 

37245  92805 

571066 

5-24 

967573 

.84 

603493 

6-09 

396507 

9 

53 

37272192704 

37299 !  92784 

57i38o 

5.24 

967522 

.85 

6o3858 

6-09 

396142 

8 

54 

571695 

5-23 

967471 

.85 

604223 

6-o8 

395777 

6 

55 

37326 192773 

572009 

5-23 

967421 

.85 

604588 

6^o8 

395412 

5 

56 

37353  92762 

572323 

5-23 

967370 

•  85! 

604953 

6.07 

390047 

4 

57 

37380  92751 

572636 

5-22 

967319 

•85  1 

6o53i7 

6.07 

394683 

3 

58 

37407 ; 92740 

572950 

5-22 

967268 

•  85' 

6o5682 

6^07 

■394318 

2 

59 

37434 j 92729 

573263 

5-21 

967217 

•  85! 

606046 

6^o6 

393954   1 

393590  0 

60 

37461 19271^ 

573575 

5-21 

967166 

.85 

606410 

6^o6 

N.  COS.  N.  Sine. 

L.  COS. 

D.l" 

L.  sine. 

J 

L.  cot. 

D.l" 

L.  tang. 

68° 

—      -   — 

— 

52 


TEIGOXOMETRICAL    FUXCTIOXS. — 22°. 


Nat.  Functions. 

LoGAKiTHMie  Functions  +  10. 

0 

|N.sine.  N.cos. 

L.  sine.  |  D.  1" 

L.COS. 

D.1" 

!  L.  tang. 

D.  1" 

L.cot 

37461  92718 

9.573575  5.21 

9.967166 

•  85 

9-606410 

6.06 

10.393590 

60 

1 

37488  92707 

573868  5-20 

967115 

.85 

606773 

6-06 

393227 

59 

2 

;375i5  92097 

574200   5-20 

967064 

•  85 

607137 

6.o5 

392863 

58 

8 

1 37342  926«6 

574512  5-19 

967013 

.85 

607500 

6-05 

392500 

57 

4 

137569  92675 

574824  !  5.19 

960961 

•  85 

607863 

6-04 

392187 
391775 

56 

5 

37595  92664 

575i36  1  5-19 

966910 

.85 

608225 

6-04 

55 

6 

37622  92653 

575447  j  5-i8 

966^59 

.85 

6o8588 

6-04 

391412 

54 

7 

37649  9'^642 

575758 

5-18 

906608 

•  85 

60S950 

6-o3 

39io5o 

53 

8 

3767b  9263 1 

576069 

5.17 

966756 

•  86 

6093 1 2 

6-03 

390688 

52 

9 

37703  92620 

5.6379 

5.17 

966705 

.86 

609674 

6-o3 

890826 

51 

10 

1 37730  92609 

576689 

D.16 

966653 

.86 

6ioo36 

6-02 

389964 

50 

11 

y77'57;92598~ 

9.576999 

5.16 

9-960002 

.86 

9'6io397 

6-02 

10-389603 

49 

12 

37784 '92387 

D77309 

5.16 

966550 

.86 

610759 

6-02 

389241 

48 

13 

3781 1  92576 

5776i« 

5.i5 

966499 

.86 

611120 

6.01 

388880 

47 

14 

37838  92505 

577927 

5.i5 

966447 

•  86 

611480 

6^01 

368520 

46 

15 

1 37865 :  92554 

578236 

5.14 

966395 

.86 

611841 

6-01 

388159 

45 

16 

137892  192543 

578545 

5.14 

966344 

.86 

612201 

6.00 

387799 

44 

17 

137919 ; 92532 

578853  ,  5-i3 

906292 

.86 

6i256i 

6-00 

387439 

43 

k 

18 

! 37946  '92521 

579162 

5-13 

966240 

.86 

•  612921 

6-00 

387079 

42 

19 

1  37973  925io 

579470 

5.i3 

966168 

.86 

613281 

5.99 

l^l^ 

41 

20 

1  37999  ■  92499 

579777 

5.12 

966186 

.86 

6i364i 

5.99 

40 

21 

138026(92488 

9-58oo«5 

5.12 

9.960065 

.87 

9.614000 

5^98 

10-386000 

39 

22 

|38o53  92477 

580392 

5.11 

966033 

.87 

614359 

5.98 

385641 

38 

23 

! 38080  92466 

580699 

5-11 

965981 

.87 

614718 

5-98 

385262 

37 

24 

1 38107  92455 

5Sioo5 

5.11 

965928 

.87 

6i5o77 

5.97 

334923 

36 

\ 

25 

j38i34  92444 

58i3i2 

5.10 

965676 

.87 

615435 

5.97 

384365 

35 

X 

26 

38i6i  92432 

581618 

5-10 

965S24 

.87 

615793 

5.97 

384207 

34 

27 

38i&8 '92421 

5S1924 

5.09 

965772 

.87 

6i6i5i 

5.96 

383649  '  33  I 

28 

!382i5  92410 

582229 

3-09 

965720 

.87 

6i65o9 

5-96 

333491 

32 

29 

38241  192399 

582535 

5.09 

965668 

•87 

616867 

5^96 

383 1 33 

31 

30 
31 

38208  '923^8 

582840 

D.o8 

960615 

•87 

617224 

5.95 

382776 

30 

38295  92377 

9.583145 

5.08 

9.965563 

.87 

9.6175S2 

5-95 

10.382418 

29 

82  1138322  192366 

583449 

5.07 

9655x1 

.87 

617939 

5.95 

382061 

28 

33  38349  92355 

583754 

D.07 

963458 

.8t 

618295 

5-94 

381705 

27 

34  ;.  38376  92343 

584058 

5-06 

965406 

•87 

6i8652 

5-94 

38 I 348 

26 

35  1:  38403  92332 

584361 

5.06 

965353 

.88 

619008 

5.94 

38099a 

25 

36  1  38430  92321 

584665 

5.06 

965301 

.83 

619364 

5.93 

380636 

24 

37  li  38456  92310 

584968 

5.o5 

965248 

.88 

619721 

5.93 

380279  i  23 

38 

i 38483 ,92299 

585272 

5.o5 

965195 

.88 

620076 

5-93 

379924  1  22 

39 

385io  922M7 

585574 

5.04 

965143 

.88 

620432 

5-92 

379568  1  21 

40 

,38537  '92276 

585877 

5-04 

965090 

•  88 

620787 

5-92 

379213  1  20 

41  1138564  192265 

9-586179 

5-03 

9.965037 

.88 

9.621142 

5.92 

10-378858  i  19 

42  I  38591 '92254 

586482 

5.o3 

964984 

.83 

621497 

5.91 

3785o3  j  18 

43  38617  1 92243 

5^783 

5-03 

964931 

•  88 

621852 

5.9. 

378148  1  17 

44  38644  92231 

587085 

5-02 

964879 

•88 

622207 

5-90 

377793  1  16 

45  38671  92220 

^2''^^5 

5.02 

964826 

•88 

622561 

5-90 

877489  1  15 

46  !t  38698  92209 

587688 

5-01 

964773 

•  88 

622915 

5-90 
5.89 

377065 

14 

47  ij  38725  192198 

587989 

5.01 

964719 

•88 

628269 
623623 

376781 

13 

48  1138752  1 92186 

588289 

5.01 

964666 

.89 

5-89 

376877 

12 

49 

38778  1 92 1 75 

588090 

5-00 

964613 

•  89 

623976 

5^8^ 

376024 

11 

50 
51 

388o5  92164 

588S90 

5.00 

964560 

•  89 

624330 

375670 

10 

38832  192152 

9.589190 

4-99 

9-964507 

•  89 

9-624683 

5.88 

10-375317 

9 

52 

38859  92141 

589489 

4-99 

964454 

•  89 

625o36 

5.88 

374964 

8 

53 

38886  92i3o 

589789 

4-99 

964400 

.89 

625388 

5.87 

374612 

7 

54  ,138912  92119 

590088 

4-98 

964347 

.89 

625741 

5.87 

374259 

6 

55  138939 '92107 

590387 

4.98 

964294 

.89 

626093 

5.87 

878907 

5 

56  !|  38966  92096 

590686 

4-97 

964240 

.89 

626445 

5.86 

373555 

4 

57  38993  '92085 

590984 

4-97 

964187 

.89 

626797 

5-86 

878203 

3 

58  ii  39020  ;  92073 

591282 

4-97 

964133 

.89 

627149 

5-86 

372851 

2 

59  ii  39046  ■  92062 

591580 

4-96 

964080 

.89 

627501 

5.85 

372499 
372148 

1 

60  i!  39073  :92o5o 

591878 

4.96 

964026 

.89 

627852 

5.85 

0 

IN.  COS. N. sine. 

L.  COS. 

D.l" 

.  L.  sine.  | 

L.  cot. 

D.  1" 

L.  tang. 

' 

67^                         1 

TRIGOJTOMETRICAL   FUNCTIONS. — 23" 


53 


Nat.  Functions. 

Logarithmic  Functions  +  10.              | 

. 

N.8ine.N.  cos.! 

L.  sine. 

D.  1" 

L.  COS. 

D.I" 

L.  tang. 

D.  1" 

L.  cot. 

0 

39073 

92o5o 

9-591878 

4-96 

9-964026 

-89 

9-627832 

5-85 

10-372148 

60 

1 

39100 

92039 

592176 

4-95 

963972 

.89! 

628203 

5-85 

371797 

59 

2 

39127 

92028 

592473 

4-95 

963919 

-89; 

628554 

5-85 

371446 

58 

8 

39153 

92016 

592770 

4-95 

963865 

.90  1 

628905 

5-84 

371093 

57 

4 

39180 

92005 

593067 

4-94 

963811 

-901 

629255 

5-84 

370745 

56 

5 

39207 

9 '994 

593363 

4-94 

963757 

-90 

629606 

5-83 

370394 

55 

6 

39234 

91982 

593659 

4-93 

963704 

-90 

629956 

5-83 

370044 

54 

7 

39260 

91971 

593955 

4-93 

96365o  -90] 

63o3o6 

5-83 

369694 

53 

8 

39287 

91959 

594251 

4-93 

963396 

-90  j 

63o656 

5-83 

369344 

52 

9 

39314 

91948 

594347 

4-92 

963542 

-90 

63ioo5 

5-82 

368995 

51 

10 

39341 

91936 

594842 

4-92 

963488 

•90  1 

63x355 

5-82 

368645 

50 

11 

39367 

91925 

9.595137 

4-91 

9-963434 

.90 

9.631704 

5-82 

10-368296 

49 

12 

39394 

91914 

595432 

4-91 

963379 

-90 

632053 

5-81 

367947 

48 

13 

39421 

91902 

593727 

4-91 

963323 

.90 

632401 

5-81 

367399 

47 

14 

39448 

91891 

596021 

4-90 

963271 

.90 

632750 

5-81 

367250 

46 

15 

39474 

91879 

596315 

4.90 

963217 

.90 

633098 

5-80 

366902 

45 

16 

39301 

91868 

596609 

4.89 

963163 

-90 

633447 

5-80 

366553 

44 

17 

39028 

91856 

596903 

4.89 

963108 

-91 

633795 

5-8o 

366203 

43 

18 

39555 

91845 

597196 

4-8? 

963o54 

-91 

634143 

5.79 

365857 

42 

19 

3938. 

91833 

597490 

962999 

-91 

634490 

5.79 

365510 

4l 

20 

39608 

91822 

597783 

4-88 

962945 

-91 

634838 

5-79 

365i62 

40 

21 

,39635 

91810 

9-598073 

4-87 

9-962890 

-91 

9.635i85 

5-78 

io-3648i5 

39 

22 

39661 

91799 

598368 

4-87 

962836 

-91 

635532 

5-78 

364468 

38 

23 

39688 

91787 

598660 

4-87 

962781 

-91 

635879 

5-78 

364121 

37 

24 

397i5 

91775 

598952 

4-86^ 

962727 

•91 

636226 

5-77 

363774 

36 

25 

1 39741 

91764 

599244 

4-86 

962672 

•91 

636572 

5-77 

363428 

35 

26 

; 39768 

91731 

599536 

4-85 

962617 

•91 

636919 

5-77 

363o8i 

34- 

27 

39795 

91741 

599827 

4-85 

962562 

.91 

637263 

5-77 

362735 

33 

28 

39822:91729 

600118 

4-85 

962308 

-91 

637611 

5-76 

362389 

32 

29 

39848191718 

600409 

4.84 

962453 

-91 

637956 

5-76 

362044 

31 

30 

139875 19,706 

600700 

4.84 

962398 

-92 

638302- 

5-76 

361698 

30» 

31 

139902191694 

9.600990 

4.84 

9-962343 

-92 

9-638647 

5-75 

10-361353 

29 

32 

139928191683 

601280 

4-83 

962288 

•92 

638992 

5-75 

361008 

28 

33 

39955191671 

601570 

4-83 

962233 

-92 

639337 

5-75 

36o663 

27 

34 

39982  91660 

601860 

4-82 

962178 

•92 

639682 

5-74 

36o3i8 

26 

85 

40008  91648 

602 i5o 

4-82 

962123 

.92 

640027 

5-74 

339973 

25 

36 

40035  91636 

602439 

4-82 

962067 

.92 

640371 

5-74 

359629 

24 

37 

40062 1 91625 

602728 

4-81 

962012 

•92 

640716 

5.73 

359284 
358940 

23. 

38 

40088' 91613 

6o3oi7 

4-8i 

961957 

-92 

641060 

5-73 

22 

39 

40II5J9160I 

6o33o5 

4-8i 

961902 

-92 

641404 

5-73 

358396 

21 

40 

40141 

91090 

603594 

4-8o 

961846 

-92 

641747 

5.72 

358253 

20 

41 

40168 

91578 

9-6o3882 

4-8o 

9-961791 

-92 

9-642091 

5-72 

10-337909 

19 

42 

40195 

91 566 

604170 

4-79 

961735 

-92 

642434 

5-72 

337366 

18 

43 

40221 

9i555 

604457 
604745 

4-79 

961680 

-92 

642777 

5.72 

357223 

17 

44 

40248  I  91543 

4-79 
4-78 

961624 

-93 

643120 

5-71 

356880 

16 

45 

40275  9i53i 

6o5o32 

ll'^.'^l 

-93 

643463 

5-71 

356537 

15 

46 

4o3oi  9i5i9 

6o53i9 

4.78 

.93 

643806 

5-71 

356194 

14 

47 

4o328  9i5o8 

603606 

4-78 

961459 

-93 

644148 

5-70 

355852 

13 

48 

4o355  1 91496 

605892 

4-77 

961402 

•93 

644490 

5-70 

355510 

12 

49 

4o38i  91484 

606179 

4-77 

961346 

.93 

644832 

5-70 

355i68 

11 

50' 

40408! 91472 

606465 

4-76 

961290 

.93 

645174 

5-69 

354826 

10 

51 

40434 [91461 

9-606751 

4-76 

9.961235 

•93 

9-645316 

5-69 

10-354484  1  9  1 

52 

40461 191449 

607036 

4-76 

961 179 

-93 

645857 

5-69 

334143 

8 

53 

40488 

9>437 

607322 

4-75 

961123 

-93 

646199 

5-69 

353801 

7 

54 

40514 

91425 

607607 

4-75 

961067 

.93 

646340 

5-68 

353460 

6 

55 

4o54i 

91414 

607892 

4-74 

9610'! 

-93 

646881 

5-68 

353119 
352778 

5 

56 

40567 

91402 

608177 

4-74 

960935 

-93 

647222 

5-68 

4 

57 

40594191390 

608461 

4-74 

960899 
960843 

-93 

647562 

5.67 

332433 

3 

58 

40621191378 

608745 

4-73 

-94 

647903 

5-67 

352097 

2 

59 

40647  9 '366 

609029 

4-73 

960786 

.94 

648243 

5-67 

351757 

1 

60 

40674  91355 

609313 

4-73 

960730 

•94 

648583 

5-66 

351417 

0 

IN.  COS.  jN.  sine 

L.  COS.  1  D.  1" 

L.  sine. 

L.  cot. 

D.l" 

L.  tang. 

* 

66°                         1 

54: 


TRIGONOMETRICAL   FL'NCTIOXS.— 24°. 


Nat.  Functions. 

Logarithmic  Functions  +  10. 

' 

N. sine. 'n.  COS.! 

Ksine.  1 

D.  1" 

L.  COS. 

D.I" 

L.  tang.  1 

Dl." 

L.  cot 

0  ! 40674! 91355 

1  :  40700  1  91843 

2  j40727!9iJ3i 
8  140753191819 

4  l|  40780  1 91807 

5  140806191295 

6  40833  j  91 283 

7  :  40860  91272 

8  ;,  40886  191 260 

9  1140913191248 
10  1  40939  91286 

9.609813 
609J57 
609S30 
610164 
610447 
610729 
611012 
611294 
611576 
6ii858 
612140 

4-73 
4-72 
4-72 
4-72 
4-71 
4-7' 
4-70 
4-70 
4-70 
4.69 
4.69 

9-960780 
960674 

•  960618 
960561 
960303 
960448 
960892 
960835 
960279 
960222 
960165 

•94 
.94 
.94 
•94 
•94 
.94 

•94 
.94 
•94 
.94 
•94 

9.643588 
648928 
649263 
649602 
649942 
65o28i 
65o62o 
630959 
65 1 297 
65i686 
631974 

5.66 
5.66 
5.66 
5.66 
5.65 
5-65 
5.65 
5.64 
5-64 
5-64 
5-63 

io.85i4i7 
351077 
350787 
350398 
350033 

349719 
349380 
349041 
348708 
348364 
348026 

60 
59 
58 
57 
56 
55 
54 
53 
52 
51 
50 
49 
43 
47 
46 
45 
44 
43 
42 
41 
40 

11  1 

12 

13 

14 

15 

16 

17 

13 

19 

20 

40966,91224 
40992  91212 
41019  91200 
41043  91188 
41072I91176 
41098 1 91 164 
4II25  91152 
4ii5i  91140 
41178  91128 
41204  91116 

9-612421 
612702 
612983 
618264 
613545 
618825 
6i4io5 
614335 
614665 
614944 

4-69 
4-68 
4-68 
4-67 
4-67 
4-67 
4-66 
4-66 
4-66 
4-65 

9-960109 
960052 

l^l 

059882 
959835 
959768 
959711 
959634 
959596 

•9? 
.93 
-93 
.93 
.93 

.95 

.93 

.93 

9.652812 
652650 
652988 
658826 
653663 
654000 
654337 
654674 
65501 1 
655843 

5.63 
5.63 
5.63 
5.62 
5.62 
5.62 
5.61 
5-61 
5-61 
5-61 

10.347688 
347350 
347012 
346674 
346887 
346000 
345663 
345326 

344989 
344652 

21  1 

22 

23 

24 

25 

26 

27 

23 

29 

SO 

41281  91104 
41257  91092 
41284  91080 
41810  91068 
4i387  9io56 

■41863  91044 
41890; 91082 
41416  91020 

'41443  91008 
41469^90996 

9.615228 
6i55o2 
615781 
616060 
616888 
616616 
616894 
617172 
617430 
617727 

4-65 
4-65 
4.64 
4-64 
4.64 
4-68 
4-63 
4-62 
4-62 
4-62 

9.959539 
959482 
959425 
959868 
959810 
959253 

95908. 

939028 

.93 
-93 
-95 

.96 
.96 
.96 
.96 
.96 
.96 

9.655684 
656020 
656856 
656692 
657028 
657864 
65i699 
658o84 
658369 
658704 

5.60 
5.60 
5-60 
5-59 

5-58 
5-58 

10.344816 
343980 
343644 
343308 
342972 
342636 
842801 
341966 
841681 
341296 

89 
88 
87 
86 
85 
84 
83 
32 
31 
30 

31 
32 
33 
34 
35 
36 
87 
33 
8'J 
40 

14149^ -90984 
4i522  90972 
41549 '90960 

41575  90948 

; 41602  90986 
141628  90924 

4i655  90911 
,41681  90899 

41707:90887 
'41734  90875 

9-6i8oo4 
618281 
6i8558 
618884 
619110 
619886 
619662 
619988 
620218 
620488 

4-6i 
4-6i 
4-6i 
4-6o 
4-6o 
4-6o 
4-59 
4-59 
4-59 
4-58 

9-958965 

958^08 
958850 
958792 
938734 
958677 
958619 
■  938561 
958303 
?58445 

.96 
-96 
.96 
.96 

.96 
-96 
•97 
•97 

9.659089 
1   659878 
659708 
i   660042 
660876 
660710 
661043 
661877 
661710 
662043 

5-58 
5.57 

ni 

5.56 
5.56 
5.55 
5.55 

10-840961 
340627 
340292 
389958 
339624 

XT, 

388628 
388290 
887957 

29 
28 
27 
26 
25 
24 
23 
22 
21 
20 

41 
42 
43 
44 
45 
46 
47 
48 
49 
50 

';  41760  90868 
i 41787 '90851 
41818:90889 
41840  90826 
41866  90814 
41892 190802 
41919  90790 
41943:90778 
41972 '90766 
,41998190753 

9.620763 
621088 
621818 
621587 
621861 
622185 
622409 
622682 
622956 
628229 

1  4-58 
4-57 

14.37 

1  4-57 
4-56 
4-56 
4-56 
4-55 
4-55 

1  4-55 

9.958887 
958829 
958271 
958213 
958 1 54 

958038 
957979 
957921 
937868 

•97 
•97 
•97 
•97 
•97 
•97 
•97 
•97 
•97 
•97 

j  9.662876 
662709 
668042 
663375 
668707 
664089 
664871 
664703 
i   665o35 
i   665366 

5-55 
5.54 
5.54 
5.54 
5-54 
5-58 
5-53 
5-53 
5.53 
5-52 

10-337624 
337291 
336958 
336625 
386293 
335961 
335629 

334684 

19 
18 
17 
]6 
15 
14 
13 
12 
11 
10 

51 
52 
53 
54 
55 
56 
57 
5S 
59 
60 

142024  90741 
I42051 190729 
j 42077 j 907 17 
142104190704 
1 42180  90692 
142156:90680 
142188:90668 

j  4^209  !  90655 
142285 '90648 
142262:90681 

g-6285o2 
628774 
624047 
624819 
624591 
624863 
625i35 
625406 
625677 
625948 

1  4-54 

1  4-54 

I.4-54 

4-53 

4-53 

!  4-53 

4.52 

4-32 

4-52 

4.51 

9.957804 
957746 

957628 
I   937570 
9575.1 
937432 
937898 
937335 
937276 

t 

.98 
.98 
-93 
.98 
.98 
.98 

1  9.665697 
J   666029 
666860 
666691 
!   667021 
1   667852 
1   667682 

!     668013 

1   668343 
668672 

5-52 
5-52 
5.5i 
5-51 
5.51 
5-5i 
5-5o 
5-50 
5.5o 
5.50 

10-884808 
388971 
333640 
338809 
332979 
382648 
882818 
381987 
381657 
381828 

9 
8 
7 
6 
5 
4 
8 
2 
1 
0 

JN.  COS.  N.  sine 

L.  COS.  1  D.  1"  J  L.  sine. 

1   L.  cot. 

D.l" 

L.  tang. 

1  ' 

65°                         1 

TRIGOl^OMETRICAL   FUXCTIOXS. — 25= 


55 


Nat.  Functions. 

Logarithmic  Functions  +  10. 

0 

N.sine. 

N.  COS. 

L.  sine. 

D.  1" 

L.  COS. 

DA" 

L.  tang. 

D  1" 

L.  cot. 

42262 

906:31 

9-625948 

4-5i 

9.957276 

.98 

9.668673 

5.5o 

io.33i327 

60 

1 

42288 

90618 

626219 

4-5i 

937217 

.98 

669002 

5-49 

330998 

69 

2 

423i5 

90606 

626490 

4-5i 

957158 

.98 

669332 

5-49 

33o668 

53 

8 

42341 

90594 

626760 

4-5o 

937099 

.98 

669661 

5-49 

330339 

57 

4 

42867 

9o582 

627030 

4-30 

937040 

.98 

669991 

5-48 

330009 

56 

5 

42394 

90569 

627300 

4-5o 

956981 

.98 

670320 

5.48 

329680 

55 

6 

42420 

90557 

627570 

4.49 

936921 

•99 

670649 

5-48 

329351 

54 

7 

42446 

90545 

627840 

4-49 

936862 

•99 

670977 

5.48 

329023 

53 

8 

42473 ! 90532 

628109 

4-49 

956803 

•99 

671306 

5-47 

328694 

52 

9 

42499 

90520 

628378 

4-48 

956744 

•99 

671634 

5-47 

328366 

51 

10 

42323 

9o5o7 

628647  1  4-48 

956684 

•99 

671963 

5.47 

328037 

50 

11 

42332 

90493 

9.628916 

4-47 

9-956625 

•99 

9.672291 

5-47 

10-327709 

49 

12 

42378 

90483 

629185 

4-47 

956566 

•99 

672619 

5.46 

327381 

43 

13 

42604 

90470 

629453 

4-47 

9565o6 

•99 

672947 

5-46 

327053 

47 

14 

4263 1 

90458 

629721 

4.46 

956447 

•99 

673274 

5-46 

326726 

46 

15 

42637 

90446 

6299S9 

4-46 

956387 

•99 

673602 

5-46 

326398 

45 

16 

42683 

90433 

630237 

4.46 

956327 

•99 

673929 

5.45 

326071 

44 

17 

42709 

90421 

63o524 

4.46 

956268 

•99 

674257 

5-43 

325743 

43 

13 

42736 

90408 

630792 

4-45 

956208 

1. 00 

674584 

5.43 

325416 

42 

19 

42762 ] 90396 

63 1039 

4-43 

956148 

1-00 

674910 

5-44 

325090 

41 

20 

42788  1  90383 

63i326 

4-45 

956089 

I -00 

675237 

5-44 

324-63 

40 

21 

42813190371 

9.631593 

4-44 

0-956029 

I -00 

9-675564 

5-44 

10.324436 

3y 

22 

42841  90338 

631859 

4.44 

955969 

I -00 

675890 

5.44 

324110 

33 

23 

42867 1 90346 

632123 

4-44 

953909 

1. 00 

676216 

5-43 

323784 

37 

24 

42894  90334 

632392 

4-43 

955849 

1. 00 

676543 

5.43 

323457 

36 

25 

42920  9032 1 

6326j8 

4-43 

955789 

1-00 

676809 

5.43 

323i3i 

35 

26 

42946  9o3oq 

632923 

4-43 

955729 

I- 00 

677194 

5-43 

322806 

34 

27 

42972  90296 

633189 

4-42 

955669 

i-oo! 

677320 

5.42 

322480 

33 

28 

42999  90284 

633454 

4-42 

935609 

1-00 

677846 
678171 

5-42 

322154 

32 

29 

43o23  90271 

633719 

4-42 

955548 

1. 00 

5.42 

321829 

31 

30 

43o5i 1 90259 

633984 

4-41 

955488 

1. 00 

678496 

5.42 

32i5o4 

30 

31 

43077  90246 

9. 634249 

4-41 

9.955428 

1.01 

9.678821 

5-41 

10.321179 

29 

32 

43 104  90233 

634314 

4.40 

955368 

I-OI 

679146 

5-41 

320854 

23 

33 

43i3o  90221 

634778 

4.40 

955307 

I  -01 

679471 

5.41 

320520 
320205 

27 

34 

43 1 56  90208 

635042 

4.40 

955247 

I  -01 

5.41 

26 

35 

43182 : 90196 

6353o6 

4-39 

955186 

I  -01 

680120 

5.40 

319880 

25 

36 

43209  90183 

635570 

4-39 

955126 

I  -01 

680444 

5-40 

319556 

24 

37 

43235  90171 

635834 

4-39 

955o65 

I-OI 

680768 

5.40 

i3i9232 

23 

33 

43261 |90i58 

636097 

4-38 

955oo5 

I-Ol 

681092 

5.40 

318908 

22 

39 

43287  I  90146 

636360 

4-38 

954944 

I  .01 

681416 

5-39 

3 18584 

21 

40 
41 

43313! 90133 

636623 

4-38 

954883 

I-Ol 

681740 

5.3^ 

318260 

20 

43340 1 90120 

9.636886 

4-37 

9.954823 

1.01 

g. 682063 

5-39 

10-317937 

ly 

42 

43366  90108 

637148 

4-37 

954762 

I-Ol 

682387 

5-39 

317613 

18 

43 

43392  90095 

63741 1 

4-37 

954701 

l-Ol 

682710 

5-38 

317290 

17 

44 

43418  900S2 

637673 

4-37 
4-36 

954640 

I-Ol 

683o33 

5-33 

316967 

16 

45 

43445  90070 

637935 

954379 

£-01 

683356 

5-38 

3 16644 

15 

46 

43471 

90037 

638197 

4-36 

954518 

1.02 

683679 

5-38 

3i632i 

14 

47 

43497 

90045 

638458 

4-36 

954457 

1.02 

684001 

5.37 

3 15999 

13 

48 

43323 

90032 

638720 

4-35 

954396 

1-02 

684324 

5.37 

3 15676 

12 

49 

f4 

90019 

638981 

4-35 

954335 

1-02 

684646 

5-37 

3 15354 

11 

^50 

90007 

639242 

4-35 

954274 

1-02 

684968 

5.37 

3i5o32 

10 

51 

43602  ;  89994 

9.639503 

4-34 

9.954213 

iT^ 

9-685290 

5-36 

io-3i47'o 

9 

52 

436281899^1 

639764 

4-34 

954152 

1-02 

685612 

5-36 

314388 

8 

53 

43634 

89968 

640024 

4.34 

954090 

1-02 

685934 

5-36 

3 1 4066 

7 

54 

43680 

89956 

640284 

4-33 

954029 

1-02 

686255. 

5-36 

3i3745 

6 

65 

43706 

89943 

640544 

4-33 

953968 

1.02 

686577 

5-35 

3i3423 

5 

56 

43733 

89930 

640804 

4-33 

953906 

1-02 

686898 

5-35 

3i3io2 

4 

57 

43759189918 

641064 

4-32 

953845 

1.02 

687219 

5-35 

812781 

8 

58 

43783 ; 89905 

641324 

4-32 

953783 

I.02I 

687540 

5.35 

3 1 2460 

2 

69 

438 n  1 89892 

64 1 583 

4-32 

953722 

i.o3 

687861 

5.34 

3i2i39 
3ii8i8 

1 

60 

43837 1 89879 

641842 

4-3i 

953660 

I -03 

688182 

5.34 

0 

l!  N.  COS.  N.  sine. 

L.  COS. 

D.l" 

L.  sine.     1 

L.cot 

D.l" 

L.tang.    '  1 

64°                           1 

56 


TRIGONOMETRICAL    FUXCTIOXS.— 26°. 


Nat.  FuKcnoxs. 

Logarithmic  Functions  •+•  10. 

'  ,X.sme.;N.  COS. 

L.  sine. 

D.  1" 

L.  cos. 

D.1-; 

L.  tang. 

D.l" 

L.  cot 

0 

43837 

89879 

9.641842 

4 

•31 

9.953660 

i.o3 

9.688182 

5.34 

io-3ii8i8 

60 

1 

43S63 

89^7 

642101 

4 

3i 

953599 

i.o3 

688502 

5.34 

311498 

59 

2 

43839 

89S54 

642360 

4 

3i 

953537 

i.o3, 

688823 

5.34 

311177 

58 

8 

43916 

8984. 

642618 

4 

3o 

953475 

103 

689143 

5-33 

310857 

57 

4 

43942 

8ob28 

642877 

4 

3o 

953413 

i-o3 

689463 

5.33 

3 10537 

56 

5 

43968  89S I 6 

643 1 35 

4 

3o 

953352 

1-03 

689783 

5-33 

310217 

55 

6 

43994  S9803 

643393 

4 

3o 

953290 

i.o3 

690103 

5.33 

309897 

54 

7 

; 44020  89790 

6436DO 

4 

29 

9D3228 

i-o3 

690423 

5-33 

309377 

53 

8 

44046  89777 

643908 

4 

29 

953166 

1-03 

690742 

5-32 

309238 

52 

9 

44072 189764 

644165 

4 

29 

953104 

1-03 

691062 

5.32 

308938 

51 

10 

44098189752 

644423 

4 

28 

953042 

i.o3 

691381 

5-32 

308619 

50 

11 

44124 189739 

9.644680 

4 

28 

9-952980 

1-04 

9-691700 

5.31 

io-3o83oo 

49 

12 

44i5i 1S9726 

644936 

4 

23 

952918 

1.04 

692019 

5-31 

307981 

4S 

13 

'44177  8o7l3 

645193 

4 

27 

932803 

104 

692338 

5.3i 

307662 

47 

U  1  442o3 ' 89700 

645450 

4 

27 

952793 

1.04 

692656 

5.31 

307344 

46 

15  !  44229  S^bSi 

645706 

4 

27 

952731 

1-04 

692975 

5.31 

307025 

45 

16  4425! 

89674 

645962 

4 

26 

952669 

1-04 

693293 

5-3o 

,  306707 
3o63S8 

44 

17 

44281 

89662 

646218 

4 

26 

952606 

1-04 

693612 

5.30 

43 

13 

44307 
44333 

89649 

646474 

4 

26 

952344 

1.04 

693930 

5-3o 

306070 

42 

19 

89530 

646729 

4 

25 

952481 

1.04 

694248 

5-3o 

3o5752 

41 

20  1  44359  89623 

646984 

4 

25 

952419 

1.04 

694366 

5.29 

3o5434 

40 

21  i  44385  S9610 

9-647240 

4 

25 

9.952356 

1.04 

9.694883 

5.29 

io-3o5ii7 

8y 

22 

4441 1  89597 

647494 

4 

24 

952294 

1.04 

693201 

5.29 

304799 

3S 

23 

44437  5Q5ti4 

647749 

4 

24 

952231 

1.04 

695518 

5.29 

304482 

37 

24  ;  44464  89571 

648004 

4 

24 

952168 

1-05, 

695836 

5.29 

304164 

36 

25  1  44490  89558 

64S258 

4 

24 

952106 

1.03 

696153 

5-23 

3o3847 

35 

26  44516  S9545 

6485i2 

4 

23 

932043 

i.o5' 

696470 
696787 

5.23 

3o353o 

34 

27  ;;  44542 

89532 

64S766 

4 

23 

951980 

i.o5 

5-23 

3o32i3 

33 

28  ;!  44568 

89519 

649020 

4 

23 

931917 

i.o5 

697103 

5-23 

302897 

32 

29  1  44594 

89506 

649274 

4 

22 

931834 

1.03 

697420 

5.27 

3o258o 

31 

30  ;  44620 

89493 

649327 

4 

22 

931791 

i.o5 

697736 

5-27 

302264 

80 

31  44646 

894:50 

9-649781 

4 

22 

9.951728 

i.o5 

9-698053 

5.27 

10.301947 

2y 

32  44672 
S3  '44698 

89467 

65oo34 

4 

22 

95 I 665 

1.03 

69S369 

5-'2 

3oi63i 

23 

89454 

65o2b7 

4 

21 

951602 

i.o5 

698685 

5.26 

3oi3i5 

27 

34  1;  44724 

89441 

65o539 

4 

21 

93.539 

I  05. 

699001 

5.26 

300999 

26 

85  j  44750 

89428 

650792 

4 

21 

931476 

i-od! 

699316 

5.26 

3oo684 

25 

36  447i6 

89415 

65 1044 

4 

20 

931412 

1.03, 

699632 

5-26 

3oo368 

24 

37 

44802 

89402 

651297 

4 

20 

951349 

1.06' 

699947 

5.26 

3ooo53 

23 

38 

44828 

8o3S9 

661549 

4 

20 

951286 

1.06 

700263 

5.23 

299737 

22 

39 

44854  89376  1 

65iSoo 

4 

19 

951222 

1-06 

700578 

5.23 

299422 

21 

40  448S0  89363 

652052 

4 

•9 

951159 

1.06 

700893 

5.25 

299107 

20 
19 

41  ,  44906  89350 

9-6523o4 

4' 

19 

9.951096 

1.06 

9.701208 

5.24 

10.298792 

42  1  44932  89337 

65z555 

4 

18 

931032 

1.06 

7oi523 

5.24 

298477 

18 

43  1  44958  89324 

652806 

4 

18 

950968 

1.06 

701837 

5-24 

298163 

17 

44  ,44984  89311 

653o57 

4 

1 3 

950905 

1.06 

702152 

5.24 

297848 

16 

45  1  45oio  8929S 

6533oS 

4 

18 

95c84i 

i.o6i 

702466 

5-24 

297534 

15 

46  1  45o36  89285 

653558 

4 

17 

950778 

1.06 

702780 

5.23 

297220 

14 

47  ,  45062  89272 

6538o8 

4 

n 

950714 

|.o6 

703093 

^'3 

296905 

13 

4S  !  45o88  89239 

654o59 

4 

17 

95o65o 

i-o6 

?o1^^ 

5-23 

296591 

12 

49  i  45i 14  89245 

654309 

4 

16 

95o5.% 

1.06 

5.23 

296277 

11 

50 

'4Pi4o  89232 

654558 

4 

16 

95o522 

1.07 

704036 

5.22 

295964 

10 

51 

'45166,89219 

9-654808 

4 

16 

"9^950438" 

1-07 

9.704350 

5.22 

10-295650 

9 

52 

'45192 '89206 

655o53 

4 

16 

950394 

1-07 

704663 

5-22 

295337 
293023 

8 

53 

45218  89193 

6553o7 

4 

i5 

95o33o 

1-07 

704977 

5.22 

7 

54 

45243  89180 

655556 

4 

i5 

950266 

1.07 

705290 

5-22 

294710 

6 

55  '45269  89167 

6558o5 

4 

i5 

950202 

1.07, 

7o56o3 

5-21 

294397 

5 

56  i452o5  8qi53 

656o54 

4 

14 

9501 38 

1-07 

705916 

5-21 

294084 

4 

57 

,45321  89140 

656302 

4 

14 

950074 

1.07 

706228 

5.21 

293772 

3 

53 

145347  89127 

656551 

4 

14 

950010 

1-07 

706541 

5.21 

293459 

2 

59 

1 453-3  8qii4 

656799 

4 

i3 

949945 

1.07. 

706854 

5.21 

293146 

1 

60  ;|  45399  S9101 

657047 

4-i3 

949881 

1.07 

707166 

5-20 

292834 

0 

jl  K  COS.  X.  sine. 

L.  COS. 

D.  1" 

L.sme. 

1 

Kcot 

D.l" 

L.tang. 

~^ 

63^ 

TRIGONOMETRICAL   rUNCTI0KS.~27°. 


57 


Nat.  Functions. 

Logarithmic  Functions  +  10 

1 

' 

N.skie.'N.cos. 

L.  sine. 

D.  1" 

L.  COS. 

D.V'\ 

L.  tang. 

D 

1" 

L-cot. 

0 

'45399 '89101 

0.607047 

4-i3 

9.949881 

1.07! 

9.707I66 

5 

20 

10.292834 

60 

1 

43425  89087 

657295 

4-i3 

949816 

1-07 

707478 

5 

20 

292522 

59 

2 

: 45451 

89074 

607042 

4-12 

949752 

1-07; 

707790 

0 

20 

292210 

5S 

3 

' 45477 

89061 

657790 

4-12 

949688 

708102 

5 

20 

291898 

57 

4 

i  455o3 

89048 

608037 

4-12 

949623 

1.08 

708414 

5 

19 

291606 

56 

5 

45529 

89035 

658284 

4-12 

949558 

1.08 

708726 

5 

•9 

291274 

55 

6 

45554 

89021 

658531 

4-11 

949494 

I- 08! 

709037 

5 

19 

290963 

54 

7 

45580 

8900S 

658778 

4-11 

949429 

1.08' 

709349 

5 

19 

290661 

53 

8 

45606 

88995 

609020 

4-11 

949364. 

1.08 

709660 

5 

19 

290340 

62 

9 

45632 

889:^1 

659271 

4-10 

949300 

1.08 

70997 1 

5 

18 

290029 

51 

10 
11 

45658 
45684 

88968 

659517 

4-10 

949235 

1. 08 

710282 

5 

18 

289718 

50 

88955 

9.609763 

4-10 

9-949170 

1.08 

9.710693 

5 

18 

10.289407 

4y 

12 

45710 

88942 

660009 

4-09 

949100 

1-08 

710904 

5 

18 

289096 

48 

13 

45736 

88928 

660255 

4-09 

949040 

1.08 

711215 

5 

18 

288780 

47 

14 

45762 

88915 

66o5oi 

4.09 

948975 

1.08 

711625 

5 

17 

288475 

46 

15 

45787 

S8902 

660746 

4.09 

948910 

1.08 

711836 

5 

17 

288164 

45 

16 

4581 3 

8^888 

660991 

4-o8 

948845 

i-o8 

712146 

5 

17 

287854 

44 

17 

45839 

88875 

661236 

4-o8 

948780 

1.09 

712456 

0 

17 

287544 

43 

18 

45865 

8S062 

661481 

4-o8 

948715 

1.09 

712766 

5 

16 

287234 

42 

19 

; 45891 

8S848 

661726 

4-07 

948650 

1-09 
1.09 

713076 

5 

16 

286924 

41 

20 
21 

: 45917 

88835 

661970 

4-07 

948584 

713386 

5 

16 

286614 

40 

45942 

88822 

9.662214 

4-07 

9.948519 

1.09 

9-713696 

0" 

10 

10-286304 

oy 

22 

45968 

8S808 

662459 

4-07 

948454 

1.09 

714005 

5 

16 

285990 

83 

23 

45994 

88795 

662703 

4 -06 

948388 

1-09 

714314 

5 

i5 

286086 

87 

24 

46020 

88782 

662946 

4 -06 

948323 

1-09 
1.09 

714624 

5 

10 

'285376 
286067 
284758 

86 

25 

: 46046 

88768 

663190 

4-o6 

948207 

714933 

5 

10 

85 

26 

' 46072 

88755 

663433 

4-o5 

948192 

1-09 

710242 

5 

i5 

34 

27 

46097 

88741 

663677 

4-o5 

948126 

1.09 

7io55i 

5 

14 

284449 

33 

28 

46123 

88728 

663920 

4-o5 

948060 

1.09 

7i586o 

5 

14 

284140 

32 

29 

46149 

88715 

664163 

4-o5 

947995 

1.10 

716168 

5 

14 

283832 

81 

80 

,46175 

88701 

664406 

4.04 

9479'^9_ 

I-IO 



1. 10 

716477 

5 

14 

283523 

80 

31 

46201 

88688 

9.664648 

4.04 

9.947863 

9.716785 

T 

14 

10.28J210 

2'J 

32 

46226 

88674 

664891 

4-04 

947797 

1. 10 

717093 

5 

i3 

282907 

23 

33 

46252 

88661 

665i33 

4-o3 

947731 

I'lO 

717401 

5 

i3 

2b2099 

27 

34 

146278 

88647 

665370 

4-o3 

947665 

I. 10 

717709 

5 

i3 

2^2291 

26 

35 

1 463o4 

88634 

665617 

4-o3 

947600 

l.IO 

718017 

0 

i3 

2 -j  1983 

25 

36 

i  46330 

88620 

665859 

4-02 

947533 

I-IO 

718325 

5 

.i3 

2^1676 

24 

37 

' 46355 

88607 

666100 

4-02 

947467 

l.IO 

718633 

5 

12 

281367 

23 

38 

46381 

88093 

666342 

4-02 

947401 

I. 10 

718940 

5 

.12 

281060 

22 

39 

46407 

88580 

666583 

4-02 

947335 

1.10 

719248 

5 

•12 

280762 

21 

40 
41 

46433 

88566 

666824 

4-01 

947269 

1.10 

719555 

5 

12 

280445 

20 

j 46458 

'8S5"53" 

9.667065 

4-01 

9-947203 

l-lO 

9.719862 

5 

12 

io.28oi38 

"iy~ 

42 

146484 

88539 

667305 

4-01 

947 1 36 

I-Il 

720169 

5 

11 

279831 

18 

43 

j465io 

83526 

667546 

4-01 

947070 

I. II 

720476 

5 

11 

279624 

17 

44 

' 46536 

88512 

667786 

4-00 

947004 

I-U 

720783 

5 

II 

279217 

16 

45 

i 46561 

88499 

668027 

4-00 

946937 

1. 11 

721089 

5 

11 

278911 

15 

40 

1 46587 

88485 

668267 

4-00 

946871 

I.I  1 

721396 

5 

11 

278604 

14 

47 

46613 

88472 
88458 

6685o6 

3.99 

946804 

1 .11 

721702 

5 

10 

278298 

13 

48 

, 46639 

668746 

3.99 

946738 

1 .11 

722009 

5 

10 

277991 

12 

.49 

,  46664 

88445 

668986 

3-99 

94667 1 

1 .11 

722310 

5 

10 

277680 

11 

50 

! 46690 

88431 

669225 

3-99 

946604 

1 .11 
l.Il 

722621 

5 

10 

277379 

10 

51 

46716 

88417 

9.669464 

3-98 

9.946538 

9.722927 

5 

10 

10.277073 

y 

52 

46742 

88404 

669703 

3.98 

946471 

1. 11 

^   723232 

5 

09 

276768 

8 

53 

' 46767 

88390 

669942 

3-98 

946404 

1. 11 

723538 

5 

09 

276462 

7 

54 

46793 

88377 

670181 

3-97 

946337 

111 

723844 

5 

09 

276166 

6 

55 

46819 

88363 

670419 

3-97 

946270 

1.12 

724149 

5 

09 

275851 

5 

56 

46844 

88349 

670658 

3-97 

946203 

1.12 

724454 

5 

09 

275646 

4 

57 

46870 

88336 

670S96 

3-97 

946 1 36 

1.12 

724759 

5 

08 

276241 

3 

58 

' 46896 

8832  2 

671 i34 

3-96 

946069 

1.12 

725o65 

5 

08 

274935 

2 

59 

46921 

883o8 

671372 

3.96 

946002 

1-12 

720369 

5 

08 

274631 

1 

60 

46947 

8S295 

671609 

3-96_ 

945935 

1.12 

725674 

5 

08 

274326 

0 

N.  008.;^.  sine. 

L.  COS. 

D.l" 

L.8ine. 

L.  cot 

^ 

1" 

L.  tang. 

' 

62°                          1 

58 


TRIGOXOMETRICAL   FUNCTIOXS. — 28" 


Nat.  Functions. 

Logarithmic  Functions  +  10.              1 

' 

N.sine.  N.cos. 

L.  sine. 

D.l" 

L.  COS.   D.l" 

L.  tang. 

D.l" 

L.cot 

0 

: 46947  882^5 

9-671609 

3-96 

9-945935  I -12 

9-725674 

5-08 

10-274326 

60 

1 

46973  88281 

671847 

3-95 

945^68  1-12 

725979 

5-08 

274021 

59 

2 

'46999  88267 

672084 

3-95 

945800  1-12 

726284 

5-07 

273716 

58 

8 

47024  88234 

672321 

3-95 

945733  I -12 

726588 

5-07 

273412 

57 

4 

47o5o  88240 

672558 

3-95 

945666  I -12 

726892 

5.07 

273108 

56 

5 

47076  8S226 

672795 

3-94 

945598  I -12 

727197 

5-07 

272803 

55 

6 

47101  8S213 

673032 

3-94 

945531  I -12 

727501 

5.07 

272499 

54 

7 

'47127  88199 

673268 

3-94 

945464  ;i-i3 

727805 

5.06 

272195 

53 

8 

i 47153  88i85 

673505 

3-94 

945396  ;i.i3 

728109 

5-06 

271891 

52 

9 

47173  88172 

673741 

3-93 

945328  i.i3 

728412 

5-06 

27i5b8 

51 

10 

47204  881 58 

673977 

3-93 

945261  |i-i3 

728716 

5-06 

271284 

50 

11 

47229  88144 

9-674213 

3-93 

9-945193  '1-13 

9-729020 

5-06 

10-270980  \   49  1 

12 

47255  88i3o 

674448 

3-92 

943125  |i-i3 

729323 

5-05 

270677 

48 

13 

47281  88117 

674684 

3-92 

945o58  ii-i3 

729626 

5-o5 

270374 

47 

14 

47306  88!  o3 

674919 

3-92 

944990  i»-'3 

729929 

5-o5 

270071 

46 

15 

47332  &80S9 

675155 

3.92 

944922  |i-i3 

730233 

5-o5 

269767 

45 

16 

47358  88075 

675390 

3-91 

944854  ii-i3 

73o535 

5-o5 

269465 

44 

17 

47383  88062 

675624 

3.91 

944786  i-i3 

73o838 

5-04 

'  269162 

43 

18 

'47409  8S048 

675859 

3-91 

944718  {I -13 

731141 

5-04 

268859 

42 

19 

47434  88o34 

676094 

3-91 

944650  ]i-i3 

731444 

5-04 

268556 

41 

20 

47460  88020 

676328 

3-90 

944582  I -14 

731746 

5-04 

268254 

40 

21 

47486  88006 

9-676062 

3-90 

9-944514  |i-i4; 

9.732048 

5-04. 

10-267952 

sy 

22 

4751 1  87993 

676796 

3-90 

944446  I -14, 

732351 

5-o3 

267649 

38 

23 

47337 ; 87979 

677030 

3-90 

944377  '•14; 

732653 

5-o3 

267347 

87 

24 

47362  87965 

677264 

3-89 

944309  1-14, 

732955 

5-03 

267045 

36 

25 

47588. 87931 

677498 

3-89 

944241  1-14, 

733257 

5-o3 

266743 

35 

26 

47614  87937 

677731 

3-89 

944172  I -14: 

733558 

5-03 

266442 

34 

27 

47639  87923 

677964 

3-88 

944104  ,1-14 

733860 

5-02 

266140 

33 

2S 

47665  87909 

678197 

3.88 

944o36  1-14 

734162 

5-02 

265838 

32 

29 

47690  87896 

678430 

3-88 

943967  ji-14 

734463 

5-02 

265537 

31 

SO 

47716 

87882 
'87C.68' 

67S663 

3-88 

943899  1-14; 

734764 

5-02 

265236 

30 

81 

47741 

9-678895 

3-87 

9 -943830  i-i4i 

9-735066 

5-02 

10-264934 

29 

32 

477<J7  87854 

679128 

3-87 

943761  jl-14; 

735367 

5-02 

264633 

28 

S3 

47793  87840 

679360 

3-87 

943693  ji-iS 

735668 

5-01 

264332 

27 

34 

47818187826 

679592 

3-87 

943624  I -15, 

735969  1  5-01 

264031 

26 

35 

47844  87812 

679824 

3-86 

943555  1. 1 5, 

736269 

5-01 

263731 

25 

36 

47869  87798 

68oo56 

3-86 

943486  i.i5| 

736570 

5-01 

263430 

24 

37 

47895  87784 

680288 

3.86 

943417  >-i5, 

736871 

5-01 

263129 

23 

33 

47920  1 87770 

68o5i9 

3-85 

943348  I -15: 

737171 

5.00 

262829 

22 

39 

47946  i  87756 

680750 

3-85 

943279  I -15, 

737471 

5-00 

262529 

21 

40 

47971  '87743 

680982 

3-85 

943210  i-i5; 

737771 

5-00 

262229 

20 

41 

,47997  87729 

9-68i2i3 

3-85 

9-943141  I -151 

9-738071 

5-00 

10-261929 

19 

42 

.48022  87715 

681443 

3-84 

943072  i-i5: 

738371 

5-00 

261629 

18 

43 

48048  87701 

681674 

3-84 

943oo3  i-i5, 

738671 

4-99 

261329 

17 

44 

148073  87687 

6S1905 

3-84 

942034  I -15; 
942864  i-i5j 

738971 

4-99 

261029 

16 

45 

!  48099  87673 

682135 

3-84 

739271 

4-99 

260729 

15 

46 

; 48124  87659 

682365 

3-83 

942795  I -16, 

739370 

4-99 

260430 

14 

47 

;48i5o  87645 

682595 

3-83 

942726  I -16; 

739870 

4.99 

26oi3o 

13 

48 

48175  87631 

682825 

3-83 

942656  I -16 

740169 

4-99 

259831 

12 

49 

;  48201 ,87617 

683o55 

3-83 

942587  I -16 

740468 

4-98 

259532 

u 

50 

'48226  87603 

683284 

3-82 

942517  1-16I 

740767 

4-98 

259233 

10 

51 

,48252  87589 

9-683514 

3-82 

9-942448  1-16; 

9-741066 

4-98 

10-258934 

"T 

52 

: 48277  8757D 

683743 

3.82 

942378  I -16, 

^  741365 

4-98 

258635 

8 

53 

483o3  87561 

683972 

3-82 

942308  1-16 

741664 

4-98 

258336 

7 

54 

'48328  87546 

684201 

3-81 

942239  I -16 

741962 

4-97 

258o38 

6 

55 

48354  87532 

684430 

3-81 

942169  I -16 

742261 

4-97 

257739 

5 

56 

.48379:87518 

684658 

3-81 

942099  I -16, 

742559  j  4-97 
742858  !  4-97 

257441 

4 

57 

48405  j  87504 

684887 
685ii5 

3-8o 

942029  i-i6| 

257142 

3 

58 

48430 

87490 

3-80 

941959  i-i6 

743 1 56  j  4-97 

256844 

2 

59 

' 48456 

87476 

685343 

3-80 

941889  I -17 

743454  4-97 

256546 

1 

60 

48481 187462 

685571 

3-80 

941819  '-17 

743752  !  4-96 

256248 

0 

i  N.  COS.  N.  sine. 

L.  COS. 

D.  1" 

L,  sine. 

L.  cot.  {  D.  1" 

L.  tang. 

' 

61°                          1 

TRIGOXOMETRICAL   FL']S^CTIONS. — 29\ 


59 


Nat.  Functions. 

Logarithmic  Functions  +  10. 

/ 

N.sme.N.  cob. 

L.  sine. 

D.  1'' 

L.  COS. 

D.l" 

L.  tang. 

D.l" 

L.  cot. 

0 

48481 

87462 

9-685571 

3 

.80 

9-941819 

1-17 

9-743752 

4.96 

10-256248 

60 

1 

485o6 

87448 

685799 

3 

•79 

941749 

1-17 

744o5o 

4-96 

255950 

59 

2 

48532 

87434 

686027 

3 

•79 

941679 

1-17 

744348 

4.96 

255652 

58 

S 

48557 

87420 

686254 

3 

•79 

941609 

1-17 

744645 

4-96 

255355 

57 

4 

1 48583 

87406 

686482 

3 

79 

941539 

1-17 

744943 

4-96 

255o57 

56 

5 

4S608 

87391 

686709 

3 

.78 

941469 

1-17 

745240 

4-96 

254760 

55 

6 

48634 

87377 

686936 

3 

78 

941398 

1-17 

745538 

4-95 

254462 

54 

7 

48659  87363 

687163 

3 

78 

941328 

1-17 

745835 

4-93 

254165 

53 

8 

48684  87349 
48710  87335 

687389 

3 

78 

941258 

1-17 

746132 

4-93 

203868 

52 

9 

687616 

3 

77 

941187 

1-17 

746429 

4-95 

253571 

51 

10 

48735  87321 

687843 

3 

77 

941117 

1-17 

746726 

4-95 

253274 

50 

11 

48761  87306 

9-688069 

T 

77 

9-941046 

1-18 

9-747023 

4.94 

10-252977 

49 

12 

148786  87292 

688295 

3 

77 

940975 

1-18 

747319 

4.94 

252081 

48 

13 

488ri '87278 

6S852I 

3 

76 

940900 

1-18 

747616 

4.94 

252384 

47 

14 

48837 

87264 

688747 

3 

76 

940834 

1-18 

747913 

4-94 

252087 

46 

15 

48862 

87200 

688972 

3 

76 

940763 

1..8 

748209 

4.94 

251791 

45 

16 

48888 

87235 

689198 

3 

76 

940693 

1-18 

7485o5 

4-93 

251495 

44 

17 

48913 

87221 

689423 

3 

75 

940622 

;:;? 

748801 

4-93 

251199 

43 

18 

48938 

87207 

689648 

3 

75 

94o5oi 

749097 
749393 

4-93 

250903 

42 

19 

48964 

87193 

689873 

3 

75 

940480 

i-i8 

4-93 

250607 

41 

20 

48989 

87178 

690098 

3 

75 

940409 

1-18 

749689 

4-93 

25o3ii 

40 

21 

49014 

87164 

9-690323 

3" 

74 

9-940338 

1-18 

1.18 

9-749985 

4-93 

io-25ooi5 

39 

22 

49040 

87150 

690548 

3 

74 

940267 

750281 

4-92 

249719 

38 

23 

49065 

87136 

690772 

3 

74 

940196 

1-18 

750076 

4-92 

249424 

37 

24 

49090 

8712J 

690996 

3 

^ 

940125 

1-19 

750872 

4-92 

249128 

36 

25 

49116 

87107 

691220 

3 

73 

940054 

1-19 

751167 

4-92 

248833 

35 

26 

49141 

87093 

691444 

3 

73 

939982 

1-19 

751462 

4-92 

248538 

34 

27 

49166 

87079 

691668 

3 

73 

93991 1 

.-19 

751757 

4-92 

248243 

33 

28 

49192 

87064 

691892 

3 

72 

939840 

1-19 

752o52 

4-91 

247948 

32 

29 

49217 

87050 

692115 

3 

72 

939768 

1-19 

752347 

4-91 

247653 

31 

80 

49242 

87036 

692339 

3 

72 

939697 

1-19 

752642 

4-91 

247358 

30 

31 

49266 

87021 

9-692002 

3 

72 

9-939620 

I -19 

9-752937 

4-91 

10-247063 

29 

82 

49293 

87007 

692785 

3 

71 

939554 

1-19 

753231 

4-91 

246769 

28 

38 

49318 

86993 

693008 

3 

71 

939482 

1-19 

753526 

4-91 

246474 

27 

84 

49344 

86978 

693231 

3 

71 

939410 

1.19 

753820 

4-90 

246180 

26 

85 

49369 

86964 

693453 

3 

71 

939339 

I -19 

754115 

4-90 

245885 

25 

86 

49394 

86949 

693676 

3 

70 

939267 

1-20 

754409 
754703 

4.90 

245591 

24 

37 

49419 

86935 

693898 

3 

70 

939195 

1-20 

4-90 

245297 

23 

38 

49445 

86921 

694120 

3 

70 

939123 

1-20 

754997 

4-90 

245oo3 

22 

89 

49470 

86906 

694342 

3 

70 

939052 

1-20 

755291 

4.90 

244709 
24441 5 

21 

40 

49495 

86892 

694064 

3 

69 

938980 

1-20 

755585 

4-89 

20 

41 

49521 

86878 

9-694786 

3 

69 

9-938908 

1-20 

9-755878 

4-89 

10-244122 

19 

42 

49546 

86863 

695007 

3 

69 

938836 

1-20' 

756172 

4-89 

243828 

18 

43 

49571 

86849 

695229 

3 

69 

938763 

1-20, 

756465 

4-89 

243535 

17 

44 

49596 

86834 

695450 

3 

68 

938691 

I  -20! 

756759 

4-89 

243241 

16 

45 

49622 

86820 

695671 

3 

68 

93S619 

1-20 

757052 

4-89 

242948 

15 

46 

49647 

8680  5 

695892 

3 

68 

938547 

1-20 

757345 

4-88 

242655 

14 

47 

49672 

86791 

696113 

3 

68 

938475 

1-20! 

757638 

4-88 

242362 

13 

48 

49697 

86777 

696334 

3 

67 

938402 

I-2I 

757931 

4-88 

242069 

12 

49 

49723 

86762 

696554 

3 

67 

938330 

1-21 

758224  4-88 

241776 

11 

50 

49748 

86748 

696775 

3- 

67 

938258 

1.2I| 

758517  4-88 

241483 

10 

51 

49773 

86733 

9-696995 

3. 

67 

9-938180 

I-2IJ 

9-758810  '  4-88 

10-241190 

9 

52 

4979B 

86719 

697210 

3- 

66 

938n3 

I-2Ii 

759102 

4-87 

240898 

8 

53 

49824 

86704 

697435 

3 

66 

938040 

1-21 

759395 

4-87 

240605 

7 

54 

49849 

86690 

697654 

3 

66 

937967 

I-2I 

759687 

4-87 

24o3i3 

6 

55 

49874 

86675 

697874 

3 

66 

937895 

I-2I 

759979 

4-87 

240021 

5 

56 

49899 

86661 

698094 

3- 

65 

937822 

I-2I 

760272 

4-87 

239728 

4 

57 

49924 

86646 

698313 

3- 

65 

937749 

I-2I 

76o564 

4-87 

239436 

3 

58 

499^0 

86632 

698532 

3- 

65 

937676 

I-2I 

760856 

4-86 

239144 

2 

59 

49975 

86617 

698751 

3 

65 

937604 

I-2l| 

761 148 

4-86 

238852 

1 

60 

5oooo 

866o3 

698970 

3-64 

937531 

I-2Ij 

761439 

4-86 

238561 

0 

N.  COS.  |n.  sine. 

L.  COS. 

D.  1" 

L.  sine. 

L.cot. 

D.l' 

L.tang. 

' 

60^                        1 

60 


TRIGONOMETRICAL   FUXCTIOXS. — 30^ 


Nat.  FUKCTION3. 

Logarithmic  Functions  +  10.              1 

0 

|N.8liie.N.co8. 

L.  sine. 

D.  1" 

L.  COS. 

D.I" 

1  Ltang. 

Dl." 

L.  cot. 

1  50000  866o3 

9-698970 

3.64 

9.937531 

1.21 

1  9-761439 

4-86 

!io.23856i 

60 

1 

;  50025  86588 

699189 

3-64 

937458 

122 

761731 

4-86 

238269 

59 

2 

'j5oo5o  865i3 

699407 

3.64 

937385 

1-22 

762023 

4-86 

237977 

58 

s 

.50076  86559 

699626 

3-64 

937312 

1-22 

762314 

4-86 

237686 

57 

4 

i5oioi  85544 

699844 

3-63 

93723s 

1-22 

762606 

4-85 

237394 

56 

5 

;5oi26i 86530 

700062 

3.63 

937165 

1-22 

762897 

4-85 

237103 

55 

6 

!5oi5i [86515 

700280 

3-63 

937092 

1.22 

763 1 88 

4-85 

236812 

54 

7 

50176  865oi 

70049S 

3-63 

937019 

1-22 

763479 

4-85 

236521 

53 

8 

'5o2oi  86486 

700716 

3-63 

936946 
936872 

1-22 

763770 

4-85 

236230 

52 

9 

1 50227  B6471 

700933 

3-62 

1-22 

764061 

4-85 

235930 
235648 

51 

10 

i  5o252 ' 86457 

7oii5i 

3.62 

936799 

1.22 

764352 

4.84 

50 

11 

1 50277 1 86442 

9-701363 

3.62 

9-936725 

1.22 

9-764643 

4-84 

10-235357 

49 

12 

i  5o3o2 I  86427 

701585 

3-62 

936652 

1.23 

764933 

4-84 

235067 

43 

13 

1 50327 !864i3 

701802 

3-61 

936578 

1-23 

765224 

4-84 

234776 

47 

14 

1 5o352 1 86398 

702019 

3-61 

9365o5 

1.23 

765514 

4-84 

234486 

46 

15 

!5o377 

86384 

702236 

3-61 

936431 

1-23 

7658o5 

4-84 

234195 

45 

16 

30403 

86369 

702452 

3.61 

936357 

1-23 

766095 

4-84 

233905 

44 

17 

! 50428 

86354 

702669 

3-60 

936284 

1-23 

766385 

4-83 

233613 

43 

18 

50453 

86340 

702885 

3-60 

936210 

1-23 

766675 

4-83 

233325 

42 

19 

50478 

86325 

7o3ioi 

3-6o 

936i36 

1-23, 

766965 

4-83 

233o35 

41 

'20 

00303 

863 10 

703317 

3-60 

936062 

1.23| 

767255 

4-83 

232745 

40 
8y 

21  i5o528:862o5 

9.703533 

3.59 

9.935988 

1-23. 

9-767545 

4-83 

10-232455 

22 

5o553 

862S1 

703749 

3.59 

935914 

1-23 

767834 

4-83 

232166 

88 

23 

50578 

86266 

703964 

3.59 

935840 

1-23! 

768124 

4-82 

231876 

37 

24 

5o6o3 

86251 

704179 

3.59 

935766 

1  -.24 

768413 

4-82 

23 1 587 

86 

25 

50628 

86237 

704395 

it 

935692 

1-24' 

768703 

4-82 

231297 

85 

26 

5o654 

86222 

704610 

935618 

I-24j 

768992 

4-82 

23 1008 

84 

27 

50679 

86207 

704825 

3.58 

935543 

I -24! 

769281  ■ 

4-82 

230719 

83 

28 

50704 

86192 

705040 

3.58 

935469 

I -241 

769570 

4-82 

23o43o 

82 

29 

50729 

86178 

705254 

3.58 

935395 

1-24; 

769860 

4-81 

23oi4o 

31 

80 

.50754 

86 1 63 

705469 

3.57 

935320 

1-24: 

770148 

4-8i 

229852 

80 

31 

50779 

86148 

9-705683 

3-57 

9-935246 

1-24 

9.770437 

4-81 

10 -229563 

29 

82 

30804 

86 1 33 

705898 

3.57 

935171 

1-24! 

770726 

4-81 

229274 
228985 

28 

33 

50829 

861  ig 

706112 

3.57 

935097 

1-24 

771015 

4-81 

27 

34 

5o854 

86104 

706326 

3.56 

935022 

1-24 

77i3o3 

4-8i 

22S697 
228408 

26 

35 

50879 

86089 

706539 
706753 

3-56 

934948 

1-24 

771592 

4-81 

25 

36 

50904 

86074 

3.56 

934873 

1-24 

77i8ao 

4-80 

228120 

2-1 

37 

50929 

86059 

706967 

3-56 

934798 

1-25 

772168 

4-8o 

227832 

23 

38 

50934 

86043 

707180 

3.55 

934723 

1-25 

772457 

4-8o 

227543 

22 

39 

50979 

86o3o 

707393 

3.55 

934649 

1.25' 

772745 

4-80 

227255 

21 

40 

5 1 004 

8601 5 

707606 

3.55 

934574 

1.25; 

773o33 

4-8o 

226967 

20 

41 

51029 

86000 

9.707819 

3-55 

9-934499 

I.25| 

9.773321 

4-8o 

10.226679 

"19" 

42 

5io54 

859S5 

708032 

3-54 

934424 

1.25 

773608 

4-79 

226392 

IS 

43 

51079 

85970 

708245 

3.54 

934349 

1-25 

773806 
774184 

4-79 

226104 

17 

44 

5no4 

85956 

708458 

3.34 

934274 

1-25 

4-79 

2258i6 

16 

45 

51129 

85941 

70S670 

3.54 

934199 
934123 

1.25 

774471 

4-79 

225529 

15 

46 

5ii54 

83926 

708882 

3-53 

1.25 

774759 

4-79 

225241 

14 

47 

51179 

85911 

709094 

3.53 

934048 

1.25 

775046 

4-79 

224954 

13 

48 

5 1 204 

85896 

709306 

3.53 

933973 

1-25 

775333 

4-79 
4-7» 

224667 

12 

49 

31229 

85881 

709518 

3-53 

933898 

1-26 

775621 

224379 

11 

50 

! 

51 

5i254 

85866 

709730 

3.53 

933822 

1.26J 

775908 

4-78 

224092 

10 

31279 

85851 

9-709941 

3.52 

9.933747 

1-26: 

9.776105 
7764S2 

4-78 

io.2238o5 

9 

52 

5i3o4 

85836 

710153 

3-52 

933671 

1.26' 

4-78 

2235i8 

8 

53 

5i329 

85821 

710364 

3.52 

933596 

.-26 

776769 

4-78 

223231 

7 

54 

5 1 354 

858o6 

710375 

3-52 

933520 

1-26 

777055 

4-78 

222945 

6 

55 

5i379 

85792 

710786 

3.51 

933445 

-26; 

777342 

4-78 

222658 

5 

56 

5 1 404 

85777 

710997 

3-51 

933369 
93329! 

-26' 

777628 

4-77 

222372 

4 

57 

51429 

85762 

7 1 1 208 

3.5i 

-26 

777915 

4-77 

222085 

8 

58 

51454 185747 

711419 

3.5i 

933217 

.26 

778201 

4-77 

221799 

2 

59 

51479  85732 

711629 

3.50 

933141 

•  26 

778487 

4-77 

22l5l2 

1 

60 

1 

1 

5i5o4  85717 

711839 

3.50 

933066 

.26, 

778774 

4-77 

221226 

0 

N.  COS.  N.  sine. 

L.  COS. 

D.l" 

L.8ine. 

I 

Lcot. 

D.l" 

L.  tang.   '  1 

59°                         1 

TRIGONOMETRICAL   FUNCTIONS.— 31°. 


61 


Nat.  Functions. 

LoGARiTHsiic  Functions  +  10. 

1 

0 

N.8ine.|N.  COS. 

L.  sine. 

D.  1" 

L.  COS.  |l 

w' 

L.  tins.     1 D.  1  " 

L.cot. 

5i5o4 135717 

9.711839 

3 

5o 

9.933066  'i 

•26 

9-778774 

4-77 

10-221226 

60 

1 

5i529 185702 

7i2o5o  3 

5o 

932990  !i 

•27 

779060 

4 

77 

220940 

59 

2 

5 1 554  85681 

712260  3 

5o 

932914  ji 

•27 

779346 

4 

76 

220654 

58 

3 

5i579  85672 

712469 

3 

49 

932838  I 

•27 

779632 

4 

76 

2  20368 

57 

4 

5i6o4  85657 

712679 

3 

49 

932762  1 

•27 

779918 

4 

76 

220082 

56 

5 

51628 185642 

712889 

3 

49 

932685  I 

•27 

780203 

4 

76 

219797 

55 

6 

5 1 653 

85627 

713098 

3 

49 

932609  I 

•27 

780489 

4 

76 

2I95II 

54 

7 

51678 

85612 

7i33o8 

3 

49 

932533  I 

•27 

780773  !  4 

76 

219225 

53 

8 

5 1 703 

85597 

713517 

3 

48 

932457  I 

•27 

781060  4 

76 

2 1 8940 

52 

9 

51728 

85582 

713726 

3 

48 

932380  I 

•27 

781346  4 

75 

218654 

51 

10 

51753 

85567 

713935 

3 

48 

932304  I 

:11 

781631  1  4 

jL 

218369 

60 

11 

5.778 

8555i 

9-714144 

3 

48 

9-932228  I 

•27 

9.781916  1  4 

75 

10.218084  j  49 

12 

5i8o3 

85536 

714352 

3 

47 

932i5i  ji 

•27 

782201  4 

75 

217799  1  48 

13 

51828 

85521 

714061 

3 

47 

932075  I 

-28 

782486  4 

75 

217514  [  47 

14 

5i852 

855o6 

714769 

3 

47 

931998  |i 

-28 

782771   4 

75 

217229  1  46 

15 

51877 

85491 

714978 

3 

47 

931921  |i 

-28 

783o56 

4 

73 

216944 

45 

16 

51902 

85476 

7i5i86 

3 

47 

931845  I 

.28 

783341 

4 

75 

•  216659 

44 

17 

01927 

85461 

715394 

3 

46 

931768  I 

-28 

783626 

4 

74 

.  216374 

43 

18 

51952 

85446 

7i56o2 

3 

46 

931691  I 

-28 

783910 

4 

74 

216090 

42 

19 

51977 

85431 

715809 

3 

46 

931614  I 

.28 

784195 

4 

74 

2i58o5 

41 

20 

52002 

85416 

716017 

3 

46 

931537  I 

.28 

784479 

4 

74 

21 552 1 

40 

21 

52026 

85401 

9-716224 

3 

45 

9-931460  II 

"^ 

9.784164 

4 

74 

10.215236 

89 

22 

52o5i 

85385 

716432 

3 

45 

^31383  I 

-28 

785048 

4 

74 

214932 

38 

23 

52076 

85370 

716639 

3 

45 

93i3o6  li 

•  28; 

785332 

4 

73 

214668 

37 

24 

52I0I 

85355 

716846 

3 

45 

931229  ;I 

•  29 

7856x6 

4 

73 

214384 

36 

25 

52126 

85340 

717053 

3 

45 

93ii52  li 

•29 

785900 

4 

73 

214100 

35 

26 

52i5i 

85325 

717259 

3 

44 

931075  I 

•29 

786184 

4 

73 

2i38i6  1  34  1 

27 

52175 

85310 

717466 

3 

44 

930998  1 

•  29 

786468 

4 

73 

213532 

33 

28 

52200 

85294 

717673 

3 

44 

930921  ii 

•29; 

786752 

4 

73 

213248 

32 

29 

52225 

85279 

717879 

3 

44 

930843  1 

•29 

787036 

4 

73 

2 1 2964 

31 

30 

52250 

85264 

718085 

3 
3' 

43 

930766  I 

•29 

787319  4 

72 

212681 

30 

2y 

52275 

85249 

9-718291 

43 

9-930688  I 

•29: 

9-787603  4 

72 

10.212397 

82 

52299 

85234 

718497 

3 

43 

930611  I 

•29 

787886 
788170 

4 

72 

212114 

23 

33 

52324 

852i8 

718703 

3 

43 

93o533  I 

•29' 

4 

72 

2ii83o 

27 

84 

52349 

85203 

718909 

3 

43 

930456  I 

•29; 

788453 

4 

72 

2ii547 

26 

35 

52374  l85iS8 

719H4 

3 

42 

930378  I 

•29 

788736 

4 

72 

211264 

25 

86 

52399! 85173 

719320 

3 

42 

93o3oo  I 

-3o' 

789019 

4 

72 

210981  1  24 

37 

52423 |85i57 

719525 

3 

42 

930223  I 

.30' 

789302 

4 

71 

210698  !  23 

88 

52448  85i42 

719730 

3 

42 

930145  I 

•3o' 

789585  1  4 

71 

2 1041 5 

22 

89 

52473  85137 

719935 

3 

41 

930067  1 

•  3o 

789868  1  4 

71 

2IOl32 

21 

40 

52498I85II2 

720140 

3 

41 

^92_9?89_  I 

-3o 

790i5i 
9-790433 

4 

71 

209849 

20 

41 

52522  '  85096 

9-720345 

3" 

41 

9-929911  I 

T3^ 

4 

71 

10-209567  !  19  1 

42 

52547 i85o8i 

720549 

3 

41 

929833  I 

-30 

790716  1  4 

71 

209284 

18 

43 

52572 ! 85o66 

720754 

3 

40 

929755  1 

.30 

790999  j  4 

71 

209001 

17 

44 

52597 

85o5i 

720958 

3 

40 

929677  I 

.30 

791281 

4 

71 

208719 

16 

45 

52621 

85o35 

721162 

3 

40 

929599  I 

.30 

791563 

4 

70 

208437 

15 

46 

52646 

85o2o 

721366 

3 

40 

929521  I 

.30 

791846 

4 

70 

2081 54 

14 

47 

52671 

85oo5 

721570 

3 

40 

929442  I 

.3o| 

792128  j  4 

70 

207872 

13 

48 

52696 

84989 

721774 

3 

39 

929364  I 

.3/ 

792410 

4 

70 

207590 

12 

49 

52720 

84974 

721978 

3 

39 

929286  I 

•31 

792692 

4 

70 

207308 

11 

50 

.52745 

84959 

722181 

3 

39 

929207  I 

-31 

792974 

4 

70 

207026 

10 

51 

52770  i  84943 

9-722385 

T 

39" 

9-929129  I 

.31 

9-793206  i  4 

70 

10-206744 

9 

52 

52794 ! 84928 

722588 

3 

j§ 

92oo5o  I 

.31 

793538  4 

69 

206462 

8 

53 

52819:84913 
52844  ,  84897 

722791 

3 

-3. 

793819  4 

69 

206181 

7 

54 

722994 

3 

38 

.31 

794101  4 

69 

205899 

6 

55 

52869  i  84882 

723197 

3 

38 

928815  I 

.31 

794383  4 

69 

2o56i7 

5 

56 

52893  ,  84866 

723400 

3 

38 

928736  I 

•  31 

794664  4 

69 

205336 

4 

57 

52918! 84851 

7236o3 

3 

37 

928657  I 

.31 

794945 

4 

69 

2o5o55 

3 

58 

52943  ;  84836 

7238o5 

3 

37 

928578  1 

-3.; 

795227 

4 

60 

204773 

2 

59 

52967  84820 

724007 

3 

37 

928499  I 

-31 

795508 

4 

6^ 

204492 

1 

60 

52992 1 84805 

724210 

3-37 

928420  1 

-3i 

795789 

4-68 

204211 

0 

N.  COS.  N,  sine. 

L.  COS. 

D.l" 

L.  sine. 

L.  cot 

D.l" 

L.  tang. 

' 

5§=                          1 

62 


TRIGONOMETRICAL   FUXCTIOXS. — 32^ 


Nat.  Functions. 

Logarithmic  Functioxs  +  10. 

/ 

|N.sine.|N.co3. 

L.  sine.  |  D.  1" 

L.  COS.   D.l" 

L.  tang. 

D.l" 

L.cot 

0 

; 52992  84805 

9-724210  '  3 

•37 

9-928420  1-32 

9-795789 

4-68 

10. 2042 II 

60 

1 

53017 j 84789 

724412  !  3 

•37 

928342  1-32 

796070 

4-68 

203930 

59 

2 

53o4i  84774 

724614  '  3 

•36 

928263  1-32 

796351 

4-68 

203649 
203368 

53 

3 

53o66  84759 

724816  :  3 

•36 

928183  1-32 

796632 

4.68 

57 

4 

53091  84743 

725017 

;  3 

•  36 

928104  !i-32 

796913 

4-68 

203087  56 

5 

|53n5  84728 

725219 

'   3 

-36 

928025  |1.32 

797194 

4-68 

202806  55 

6 

53140  84712 

725420 

;  3 

.35 

927946  I1.32 

797475 

4-68 

202325   54 

7 

53164.84697 

723622 

;  3 

.35 

927^67  I1.32 

797755 

4-68 

202245  53 

8 

53189:84681 

725823 

■  3 

.35^ 

927787  1-32 

798036 

4-67 

201964  1  52 

9 

153214  84666 

726024 

i  3 

.35 

927708  1-32 

798316 

4-67 

2016S4  51 

10 

53238 ' 8465o 

726225 

i  3 

35 

^  927629  [1.32 

798596 

4-67 

201404  50 

11 

53263  84635 

9.726426 

1 1 

34 

9.927^49  :i-.'^2, 

9-798877 

4-67 

10-201123 

49 

12 

'53288  84619 

•726626 

'  3 

34 

927470 

1.33' 

799137 

4-67 

200843 

4S 

13 

'53312  84604 

726827 

3 

34 

927390 

1.33! 

799437 

4-67 

200563 

47 

U 

53337  84588 

727027 

3 

34 

927310. 

1.33' 

799717 

4-67 

200283 

46 

15 

' 53361  84573 

727228 

3 

34 

927231 

1.33, 

799997 

4-66 

2oooo3 

45 

16 

'53386  84557 

727428 

3 

33 

927151 

1-33 

800277 

4.66 

199723 

44 

17 

,5341 1' 84542 

727628 

3 

33 

927071  ii-33; 

800557 

4-66 

199443 

43 

18 

1 53435  84526 

727828 

3 

33 

926991  11-33 

8oo836 

4-66 

199164 

42 

19 

,53460  84011 

728027 

3 

33 

926011  I1.33 
926831  !i.33, 

801 1 16 

4-66 

198884 

41 

20 

53484  S4495 

728227. 

3 

33 

801396 

4-66 

19S604 

40 

21 

53509  84480 

9 -.728427 

3 

32 

9-926751 

1-33 

9-801675 

4-66 

10. 198325 

39 

22  i:  53534  84464 

72S626 

3 

32 

926671 

1.33 

801955 

4-66 

198045 

S3 

23  ;  53558  S4448 

728825 

3 

32 

926591 

1.33' 

802234 

4-65 

197766 

87 

24  !  53583  84433 

729024 

3 

32 

9265ii 

1-34: 

8025 1 3 

4-65 

197487 

86 

25  53607  84417 

729223 

3 

31 

926431 

1-34' 

802792 

4-65 

197208 

85 

26  ;' 53632  84402 

729422 

3 

3i 

926351 

1-34 

803072 

4-65 

196928 

84 

27  !  53656  84386 

729621 

3 

3i 

926270 

1.34; 

8o335i 

4-65 

196649 

83 

23  i:5368i  843TO 

729820 

3 

3i 

926190 

1.34 

8o363o 

4-65 

196370 

32 

29  1]  53705  ,  84355 

730018 

3 

3o 

9261 10 

1.34 

803908 

4-65 

196092 

31 

80  !i  53730  ,  84339 

730216 

3 

3o 

926029  !i.34 

804187 

4-65 

195813 

30 

31  153754  84324 

9-73o4i5  1  3 

3o 

'■untti'di 

9-804466  4.64 

10-195534 

29 

82  I  53779 ' 84308 

73o6i3  3 

3o 

804745 

4.64 

195255 

28 

S3  53804 '84292 

730811   3 

3o 

9257S8  1 1. 34' 

8o5o23 

4.64 

194977 

27 

U 

53828  84277 

731009  1  3 

29 

925707  ii.34: 

8o53o2 

4-64 

194698 

26 

85 

53853  84261 

731206 

3 

29 

925626  ;i.34' 

8o558o 

4-64 

194420 

25 

36 

53877  84245 

•731404 
f3 1602 

3 

29 

925545  I1-35 

8o5859 

4.64 

194141 

24 

G7 

53902  84230 

3 

29 

925465  ji.35 

806137 

4.64 

193863 

23 

38 

,53926  84214 

731799 

3 

It 

925384  1.35: 

806415 

4-63 

193585 

22 

39 

,53951  84.98 

731996 

3 

9253o3  I1.35' 

806693 

4-63 

193307 

21 

411 

i 53975  84182 

732193 

3 

28 

925222  I1.35, 

806971 

4-63 

193029 

20 

41 

'54000  84167 

9.732390 

3 

28 

9-925141  .;1.35 

9-807249 

4-63 

10-192751 

19 

42 

54024  84i5i 

732587 

3 

28 

925060  1.35 

807527 

4-63 

192473 

18 

43 

54040  84135 
54073  84120 

732784 

3 

28 

924079  1-35 

807805 

4-63 

192195 

17 

44 

732980 

3 

27 

924897 

1.35, 

808083 

4-63 

191917 

16 

45 

54097184104 

733177 

3 

27 

924816 

1.35: 

8o836i 

4-63 

191639 

15 

46 

54122:84088 

733373 

3 

27 

924735 

1.36; 

8o8638 

4-62 

191362 

14 

i7 

54146 !  84072 

733569 

3. 

27 

924654 

1.36' 

808916 

4-62 

191084 

18 

43 

54171  84057 

733765 

3- 

27 

924572 

1-36; 

809193 

4-62 

190807 

12 

49 

54195  84041 

733961 

3. 

26 

924491  1-36 

809471 

4-62 

190329 

11 

50 

54220 '  84025 

734157 

3. 

26 

924409  'i.36 

809748 

4-62 

190252 

10 

51 

54244  84009 

9-734353 

"3^ 

W 

9-924328  17736 
924246  ii.36 

9.810025 

4-62 

10.189975 

9 

52 

54260  83994 
5429!  83978 

734549 

3. 

26 

8io3o2 

4-62 

189698 

8 

53 

73^744 

3. 

25 

924164  ;i.36 

8io58o 

4-62 

189420 

7 

54 

54317  83962 

734939 

3. 

25 

9240S3  1.36 

810857 

462 

188866 

6 

55 

54342  83946 

73513! 

3- 

25 

924001  ji.36 

811134 

4-61 

5 

56 

54366  8393o 

735330 

3- 

25 

923919  i.36j 

811410 

4-61 

188590 

4 

57 

54391  83015 
54415  83899 
54440  83883 

735525 

3. 

25 

923837  1.36 

811687 

4-61 

1883 1 3 

3 

58 

735719 

3. 

24 

923755  1.37 

811964 

4.61 

i88o36 

2 

59 

735914 

3. 

24 

923673  '1.37 

812241 

4-61 

187759 

1 

60 

54464  83867 

736109 

3.24 

923591  1.37 

812517 

4.61 

187483 

0 

N.  COS.  N.  sine. 

L.  COS. 

D.  1  " 

L.  sine.      ji 

L.cot.  1  D.  1"| 

L.tang. 

/ 

57-                         1 

TRIGONOMETRICAL   FUNCTIONS. — 33". 


63 


Nat.  Functions.  1 

Logarithmic  Functions  +  10. 

"' 

N.eine.l 

N.  COS. 

L.  sine. 

D.  1" 

L.COS.   D.l"|i 

L.  tang.  1  D.  1" 

L.cot. 

0 

54464  1 

83867 

9.736109 
7363o3 

3-24 

9-923591 

•37! 

9-812517 

4-6i 

0-187488 

60 

1 

54488  1 

83851 

3-24 

923509 

•37 

812794 

4-6i 

187206 

59 

2 

54513] 

83835 

786498 

8-24 

928427 

■  •87, 

818070 

4-6i 

186930 

53 

8 

54537  1  83819 

786692 

8-23 

928845 

••371 

818847 

4-60 

186653 

57 

4 

54561  838o4 

736886 

3-23 

928268 

1-37 

818628 

4-6o 

186877 

56 

5 

54586  83788 

787080 

3-23 

928181 

1.37, 

818899 
814173 

4-6o 

186101 

55 

6 

54610  83772 

787274 

3.28 

928098 

1-37' 

4-60 

185825 

54 

7 

54635 1 83756 

787467 

8-23 

928016 

1. 871 

814452 

4-60 

185548 

53 

8 

54659  ;  83740 
54683  .83724 

787661 

3-22 

922933 

1-37I 

814728 

4-6o 

185272 

52 

9  ! 

787855 

3-22 

92285i 

1.87! 

■8i5oo4 

4-6o 

184996 

51 

10! 

54708  1  83708 

788048 

3-22 

922768 

1-88! 

i.38| 

815279 

4-6o 

184721 

50 

11  1 

54732  ,  83692 

9-738241 

3-22 

9-922686 

9-815555 

4-59 

10-184445 

49 

12  1 

54756  !  83676 

788434 

3-22 

922608 

1-881 

81 583 1 

4-59 

184169 

48. 

13 

54781  1 83660 

788627 

3-21 

922520 

1-8^1 

816107 

4-59 

188898 

47 

14! 

54805  !  83645 

788820 

8-21 

922488 

i-88i 

816882 

4-59 

188618 

46 

15  1 

54829 183629 

789013 

3-21 

922855 

1-381 

8i6658 

4-59 

188342 

45 

16  ! 

54854  836i3 

789206 

3-21 

922272 

1-38; 

816933 

4-59 

188067 

44 

17 

54878 

83597 

789898 

3-21 

922189 

1-381 

817209 

4.59 

182791 

43 

18  1 

54902 

83581 

789090 

3-20 

922106 

I  •38! 

817484 

4-59 

182516 

42 

19 

54927 

83565 

789788 

3-20 

922028 

1-38' 

81^23? 

^12 

182241 

41 

20 
21 

54951 

83549 

789975 

3-20 

921940 

1-881 
1-39 

4-58 

181965 

40 

54975 

83533 

9-740167 

3-20 

9-921857 

9-8i83io 

4-58 

10-18x690 

89 

22 

54000 

83517 

740359 

3-20 

921774 

1-89 

8x8585 

4-58 

i8i4i5 

33 

23  55024 

83501 

74o55o 

8-19 

921691 

818860 

4-58 

181140 

37 

24 

55048 

83485 

740742 

8-19 

921607 

1.89 

819185 

^•^^ 

i8o865 

36 

25 

55072 

83469 
83453 

740934 

3-. 9 

921524 

1.89 

819410 

4-58 

180590 

85 

26 

55097 

741125 

8-19 

921441 

1  -39! 

819684 

4-58 

180816 

34 

27 

•55I2I 

83437 

741816 

3-19 

921857 

1.39I 

819959 

4-58 

180041 

33 

23 

,55145 

83421 

74i5o8 

8-i8 

921274 

1-391 

820284 

4-58 

179766 

82 

29 

'  55169 

83405 

741699 

3-i8 

921190 

1-89; 

82o5o8 

4.57 

179492 

31 

30 

55194 

83389 

741889 

3-18 

921107 

1-39, 

820-83 

4-57 

179217 

80 

81  ,155218 

83373 

9-742080 

3-18 

9-921028 

1-89 

9-82'io57 

4-57 

10-178948 

29 

32 

; 55242 

83356 

742271 

3-18 

XA 

1-40, 

82x882 

4-57 

178668 

28 

33 

! 55266 

83340 

742462 

3-17 

1-40 

821606 

4-57 

178894 

27 

34 

:  55291 

83324 

742652 

8-17 

920772 

I  -40 

82x880 

4.57 

178120 

26 

35 

!553i5 

833o8 

742842 

8-17 

920688 

1.40, 

822x54 

4-57 

177846 

25 

36 

i  55339 

83292 

743o33 

8-17 

920604 

1-40 

822429 

4-57 

177571 

24 

37 

. 55363 

83276 

748228 

3.17 

920520 

..40 

822708 

^1z 

177297 

23 

3S 

, 55388 

83260 

748418 

3-i6 

920486 

1-40 

822977 

4-56 

177028 

22 

89 

55412 

83244 

748602 

8-i6 

920352 

I -40! 

828250 

4-56 

X767DO 

21 

40 

1 55436 

83228 

748792 

8-i6 

920268 

1-40 

823524 

4-56 

176476 

20 

'~i\ 

1 55460 

83212 

9.748982 

8-i6 

9-920184 

1.40 

9-82-8798 

4-56 

10-176202 

19 

42 

1 55484 

83i95 

744171 

3.)6 

920099 

1-40 

824072 

4-56 

175Q28 

18 

43 

55509 

83x79 

744861 

3-i5 

920013 

1-40 

824345 

4-56 

1756D5 

17 

44 

55533 

83i63 

744550 

3-15 

919981 

1-4., 

824619 

4-56 

175381 

16 

45 

55557 

83 1 47 

744789 
744928 

8-i5 

919846 

1-41 

8248q3 

4-56 

175107 

15 

46 

55581 

83i3i 

8-i5 

919762 

1-41 

825x66 

4-56 

174884 

14 

47 

556o5 

83ii5 

745117 

3-i5 

919677 

1-41 

825439 

4-55 

174561 

13 

48 

55630 

83098 

745806 

8-14 

919598 

!i-4i 

825718 

4-55 

174287 

12 

49 

55654 

83o82 

745494 

3.14 

919508. 

1-41 

825986 

4-55 

I74bi4 

11 

90 

55678 

83o66 

745688 

3-14 

919424 

1-41 

826259 

4-55 

178741 

10 

'51 

jl  55702 

83o5o 

9-745871 

3-14 

9-919339 

|i-4i 

9-826532 

4^55 

10-178468 

9' 

52 

55726 

83o34 

746039 
746248 

3.14 

919254 

|i-4i 

826805 

4-55 

178195 

8 

&3 

55750 

83oi7 

3-i3 

919169 

1-41 

827078 

4-55 

172922 

7 

64 

55775 

83001 

746486 

3-13 

919085 

I-4I 

827351 

4-55 

172649 

6 

55 

f^ 

1 82985 

746624 

3-i3 

919000 

I-4I 

827624 

4-55 

172876 

5 

56 

82969 

746812 

8-i3 

918915 

1-42 

827897 

4-54 

172103 

4 

57  !j  55847 

82953 

746999 
747187 

3.i8 

918880 

1-42 

828170 

4-54 

171880 

8 

58  55871 

82936 

3-12 

918745 

1-42 

828442 

4-54 

171558 

2 

59  l|  55895 

82920 

747374  3-12 

918659  1-42 

828715 

4-54 

171285 

1 

60  :|  55919 

82904 

747562   3-12 

918574  ;i-42 

828987 

4-54 

171018 

0 

|i  N.  COS. 'n.  sine 

L.  COS.  1  D.  \" 

L.  sine.  1 

L.cot  , 

D.  1" 

L.  tang. 

' 

56° 

1 

64 


TRIGONOMETRICAL   FCXCTIONS. — 34° 


Nat.  Functions. 

Logarithmic  Functions  +  10.             .  1 

0 

N.sine.^N.  COS. 

L.  sine. 

D.  1" 

L.  COS.   D.l" 

L.  tang. 

D.l" 

L.  cot 

55919 

82904 

9-747562 

3 

12 

9-918574  I 

42 

9-828987 

4 

54 

10-171013 

60 

1 

55943 

82887 

747749 

3 

12 

918489  I 

42 

829260 

4 

54 

170740 

59 

2 

55968 

82871 

747936 

3 

12 

918404  1 

42 

829532 

4 

54 

170468 

58 

3 

55992 

82855 

748123 

3 

II 

9i83i8  I 

42 

829805 

4 

54 

170195 

57 

4 

|56oi6 

82839 

748310 

3 

II 

918233  1 

42 

830077 

4 

54 

169923 

56 

5 

56040 

82822 

7484Q7 

3 

II 

918147  I 

42 

83o349 

4 

53 

169651 

55 

6 

56o64 

82806 

7486S3 

3 

II 

918062  I 

42 

83o62i 

4 

53 

169379 

54 

7 

1 56o88 

82790 

748870 

3 

II 

917076  1 
917891  I 

43 

830893 

4 

53 

169107 

53 

8 

56II2 

82773 

749056 

3 

10 

43 

83ii65 

4 

53 

168835 

52 

9 

56 1 36 

82757 

749243 

3 

lO 

917805  I 

43 

8I1437 

4 

53 

168563 

51 

10 
11 

56 160 

82741 

749429 

3 

10 

917719  I 

43 

831709 

4 

53 

168291 

50 

56 1 84 

82724" 

9-74o6i5 

3 

10 

9-917634  I 

T3 

9-83i98i 

4 

53 

iO'i68oi9 

49 

12 

56?o8 

8270S 

749801 

3 

10 

917548  I 

43 

832253 

4 

53 

167747 

48 

13 

56232 

82692 

749987 

3 

09 

917462  I 

43 

832525 

4 

53 

167475 

47 

14 

56256 

82675 

750172 

3 

09 

917376  I 

43 

832796 

4 

53 

167204 

46 

15 

56280 

82659 
82643 

75o358 

3 

09 

917290  I 

43' 

833o68 

4 

52 

166932 

45 

16 

563o5 

750543 

3 

09 

917204  I 

43 

833339 

4 

52 

16666 I 

44 

17 

56329 

82626 

750729 

3 

S 

917118  I 

44 

83361 1 

4 

52 

166389 
166118 

43 

18 

56353 

82610 

750914 

3 

917032  I 

44 

833882 

4 

52 

42 

19 

56377 

82593 

751099 

3 

08 

916946  I 

44 

834154 

4 

52 

165846 

41 

20 

56401 

82577 

751284 

3 

08 

916859  I 

44 
44 

834425 

4 

52 

165575 

40 

21 

56425 

82561 

9-751469 

3 

08 

9-916773  I 

9.834696 

4 

"52 

io.i653o4 

3y 

22 

56449 

82544 

75i654 

3 

08 

916687  . 

44 

834967 

4 

52 

i65o33 

38 

23 

56473 

82528 

75i839 

3 

08 

916600  1 

44 

835238 

4 

52 

164762 

37 

24 

56497 

825ii 

752023 

3 

07 

9i65i4  I 

44 

835509 

4 

52 

164491 

36 

25 

56521 

82495 

752208 

3 

07 

916427  I 

44 

335780 

4 

5i 

164220 

35 

26 

56545 

82478 

752392 

3 

07 

916341  I 

44 

836o5i 

4 

5i 

163949 
163678 

34 

27 

56569 

82462 

■    752576 

3 

07 

916254  I 

44 

836322 

4 

5i 

33 

28 

56593 

82446 

752760 

3 

07 

916167  I 

45 

836593 

4 

5i 

163407 

32 

29 

56617 

82429 

752944 

3 

06 

916081  I 

45 

836864 

4 

5i 

i63i36 

31 

30 
31 

56641 

82413 

753128 

3 

06 

915994  I 

45 

837134 

4 

5i 

162866 

30 

56665 

82396 

9-753312 

3 

06 

9-915907  I 

% 

9-837405 

4 

5i 

10-162595 

29 

32 

56689 

82380 

753495 

3 

06 

915820  I 

837675 

4 

5i 

162325 

28 

83 

56713 

82363 

753679 

3 

06 

915733  1 

45 

837946 

4 

5i 

162054 

27 

34 

56736 

82347 

753862 

3 

o5 

915646  I 

45 

838216 

4 

5i 

161784 

26 

35 

56760 

82330 

754046 

3 

o5 

915559  I 

45 

838487 

4 

5o 

i6i5i3 

25 

36 

56784 

82314 

75ir229 

3 

o5 

915472  I 

45 

838757 

4 

5o 

161243 

24 

87 

568o8 

82297 

754412 

3 

o5 

915385  I 

45 

839027 

4 

5o 

160973 

23 

38 

56832 

82281 

754595 

3 

o5 

915297  1 

45 

839297 

4 

5o 

160703 

22 

89 

56856 

82264 

754778 

3 

04 

9i52io  I 

45 

839568 

'4 

5o 

160432 

21 

40 

5688o 

82248 

754960 

3 

04 

9i5i23  1 

46 

839838 

4 

5o 

160162 

20 

41 

56904 

8223/ 

9-755143 

3 

04 

9 -915035  1 

46 

9-840108 

4 

5o 

10-159892 

19 

42 

56928 

82214 

755326 

3 

04 

914948  I 

46 

840378 

4 

5o 

i5q622 

IS 

43 

56952 

82198 
82181 

755508 

3 

04 

914860  I 

46 

840647 

4 

5o 

159353 

17 

44 

56976 

755690 

3 

04 

914773  I 

46 

840917 

4 

49 

159083 

16 

45 

57000 

82165 

755872 

3 

o3 

914685  1 

46 

841187 

4 

49 

I6y8i3 

15 

46 

57024 

82148 

756o54 

3 

o3 

914598  I 

46 

841457 

4 

49 

158543 

14 

i7 

57047 

82132 

756236 

3 

o3 

914510  I 

46 

841726 

4 

49 

158274 

13 

43 

57071 

82115 

756418 

3 

o3 

914422  I 

46 

841996 

4 

49 

1 58004 

12 

49 

57.595 

82098 

756600 

3 

o3 

•914334  I 

46 

842266 

4 

49 

157734 

11 

50 

071 19 

82082 

756782 

3 

02 

914246  I 

il 

842535 

4 

49 

157465 

10 
'  T 

51 

157143 

82065 

9-756963 

3" 

02 

9-914158  I 

47 

9.842805 

4' 

49 

10-157195 

52 

[5ti67 

82048 

757144 

3 

02 

914070  I 

47 

843074 

4 

49 

156926 

8 

53 

157191 

82032 

757326 

3 

02 

913982  1 

47 

843343 

4 

49 

156657 
156388 

7 

54 

: 572.5 

82015 

757507 

3 

02 

913894  I 

47 

843612 

4 

49 

6 

55 

i 57238 

81999 

757688 

3 

01 

9i38o6  1 

47 

843882 

4 

48 

156118 

5 

56 

57262 

81982 

757869 

3 

01 

913718  I 

47 

8441 5i 

4 

48 

155849 

4 

57 

57286 

81965 

758o5o 

3 

01 

9i363o  I 

47 

844420 

4 

48 

1 55580 

8 

58 

:573jo 

81949 

758230 

3 

01 

913541  I 

47 

844689 

4 

48 

i553ii 

2 

59 

57334 

81932 

758411 

3 

01 

913453  I 

47 

844958 

4 

48 

155042 

1 

60 

57358 

81915 

758591 

3  01 

913365  ii 

47 

845227 

4.48 

154773 

0 

11.  COS. 

N.  sine. 

L.  COS.   D.  1" 

L.  sine.      | 

L.  cot.  !  D.  1" 

L.  tang. 

' 

55° 

TRIGONOMETRICAL   FUNCTIONS. — 55°. 


65 


Nat.  Functions. 

Logarithmic  Functions  +  10.              j 

1 

0 

•N.sine. 

N.  COS. 

L.  sine.  |  D.  1" 

L.  COS. 

D.l" 

L.  tang. 

D.l" 

L.  cot.  1    1 

57358 

81915 

9.758591 

3-01 

9.913355 

1-47 

i  9-845227 

4.48 

10.154773 

60 

1 

57381 

81899 

758772 

3.00 

913276 
913187 

|i-47 

845496 

4-48 

1 54504 

59 

2 

574o5 

81882 

758952 

3-00 

1-48 

845764 

4-48 

154236 

53 

3 

^%l 

8 1 865 

759132 

3-00 

913099 

1-48 

846033 

1  4-48 

153967 

57 

4 

81848 

759312 

3-00 

9i3oio 

'1.48 

846302 

i  4-48 

153698 

56 

5 

^lAll 

8i832 

759492 

3-00 

912922 

1-48 

846570 

1  4-47 

1 53430 

55 

6 

57501 

8i8i5 

759672 

2.99 

912833 

1.48 

846839 

4-47 

i53i6i 

54 

7 

57524 

81798 

759852 

2-99 

912744 

1.48 

847107 

4-47 

152893 

53 

8 

57548 

81782 

760031 

2-99 

912655 

1.48 

847376 

4-47 

152624 

52 

9 

57572 

81765 

7602 I I 

2-99 

912566 

1.48 

847644 

4-47 

152356 

51 

10 

575q6 

81748 

760390 

2-99 

912477 

1.48 

8479 '3 

4-47 

152087 

50 

11 

57619 

81731 

9.760569 

2.98 

9.912388 

i~48 

9.848181 

4-47 

io-i5i8i9 

4'J 

12 

57643 

81714 

760748 

2.98 

912299 

1-49 

848449 

4-47 

i5i53i 

4S 

13 

57667 

81698 
81681 

760927 

2-98 

912210 

1-4Q 

848717 

4-47 

i5i283 

47 

14 

57691 

761106 

2.98 

■912121 

1-49 

848986 

4-47 

i5ioi4 

46 

15 

57715 

81664 

761285 

2-98 

9i2o3x 

1-49 

849254 

4-47 

1 50746 

45 

16 

57738 

81647 

761464 

2-98 

9IIQ42 

911853 

1.49 

849522 

4-47 

i5o4-S 

44 

17 

57762 

8i63i 

761642 

2.97 

1.49 

849790 

4-46 

l502I0 

40 

18 

57786 

81614 

76:821 

2-97 

911763 

1.49 

85oo58 

4.46 

149942 

42 

19 

57810 

81597 

761999 

2-97 

91 1674 

1-49 

85o325 

4-46 

149673 

41 

20 

57833 

8i58o 

762177 

2-97 

9II584 

1-49 

85o593 

4.46 

149407 

4" 

21 

57857 

8 1 563 

9.762356 

2-97 

9-911495 

1-49 

9.830861 

4-46 

10.149139 

3y 

22 

57881 

81546 

762534 

2.96 

911405 

1-49 

851129 

4-46 

I48S71 

38 

23 

57904 

8i53o 

762712 

2.96 

9ii3i5 

I -So 

85i396 

4-46 

148604 

37 

24 

57928 

8i5i3 

762889 

2-96 

911226 

1-50 

85 1 664 

4-46 

148336 

36 

25 

57952 

81496 

763067 

2.96 

9ili36 

i-5o 

85i93i 

4-46 

14S069 

35 

26 

57976 

81479 

763245 

2.96 

911046 

i-5o 

852199 

4-46 

147801 

34 

27 

57999 

81462 

763422 

2.96 

910956 

I -50 

852466 

4.46 

147534 

S3 

28 

58023 

81445 

763600 

2-95 

910866 

i-5o 

852733 

4-43 

147267 

32 

29 

58047 

81428 

763777 

2.95 

910776 
910686 

1-50 

853001 

4-45 

146999 

31 

30 
31 

58070 

81412 

763954 

2.95 

1-50 

853268 

4-45 

146732 

So 

58094 

81395 

9.764131 

2.95 

9-910596 

1-50 

9-853535 

4-45 

10.146465 

2y 

32 

58ii8 

81378 

764308 

2.95 

9io5o6 

i-5o, 

853802 

4-45 

146198 

2S 

33 

58i4i 

8i36i 

764485 

2-94 

910415 

1-50 

854069 

4-45 

145931 

27 

34 

58i65 

81344 

764662 

2.94 

910825 

i.5i 

854336 

4-45 

145664 

26 

35 

58189 

8i327 

764838 

2.94 

910235 

i-5i 

8546o3 

4-45 

145397 
i45i3o 

25 

36 

58212 

8i3io 

765oi5 

2-94 

910144 

i.5i 

854870 

4-45 

24 

37 

58236 

81293 

765191 

2.94 

910004 

i-5i: 

855137 

4-45 

144S63 

23 

38 

58260 

81276 

765367 

2-94 

909963 

i.5ii 

855404 

4-43 

144396 

22 

39 

58283 

81259 

765544 

2.93 

909873 
909782 

1-51 

855671 

4-44 

144329 

21 

40 

58307 

81242 

765720 

2.93 

i.5i! 

855933 

4-44 

144062 

20 

41 

58330 

81225 

9.765896 

2.93 

9-909691 

1-51 

9-856204 

4.41 

10.143796 

ly 

42 

58354 

81208 

766072 

2.93 

909601 

i.5i 

856471 

4-44 

143329 

18 

43 

58378 

81191 

766247 

2.93 

909510 

i-5i. 

856737 

4.44 

143263 

17 

44 

58401 

81174 

766423 

2.93 

909410 
909328 

i.5i 

837004 

4.44 

142996 

16 

45 

58423 

81I57 

766598 

2-92 

1-52 

857270 

4-44 

142730 

15 

46 

58449 

81140 

766774 

2.02 

909237 

1.52 

857537 

4.44 

142463 

14 

47 

58472 

81123 

766949 

2.92 

909146 

1-52 

857803 

4.44 

142197 

13 

48 

58496 

81106 

767124 

2.92 

909055 
908964 

1-52 

858069 

4.44 

141931 

12 

49 

58519 
58543 

81089 

767300 

2-92 

1.52i 

858336 

4.44 

141664 

11 

r50 

81072 

767475 

2-91 

908S73 

1-52, 

858602 

4.43 

141398 

10 

51 

58567 

8io55 

9.767649 

2.91 

9 •908781 

1.52! 

9-858868 

4-43 

io.i4ii32 

9 

52 

58590 

8io38 

767824 

2.91 

908690 

1-52 

859134 

4.43 

140866 

8 

53 

586i4 

81021 

?a??? 

2.91 

908599 

..52| 

859400 

4.43 

140600 

7 

54 

58637 

81004 

2-91 

908507 

1.52' 

839666 

4-43 

140334 

6 

55 

58661 

80987 

768348 

2-90 

908416 

1.53 

859932 

4.43 

140068 

5 

56 

58684 

80970 

768522 

2-90 

908324 

1-53: 

860198 

4-43 

139802 

4 

57 

58708 

80953 

768697 

2.90 

908233 

1-53; 

860464 

4.43 

189536 

8 

53 

58731 

80936 

768871 

2.90 

908141 

1-53, 

860730 

4-43 

189270 

2 

59 

58755 

80919 

769045 

2-90 

908049 

1.53; 

860995 

4.43 

189005 
188789 

1 

60 

58779 

80902 

769219 

2.90 

907958 

1.5:5 

861261 

4-43 

(t 

N.  COS. 

N.sine. 

L.  COS. 

D.l" 

L.  sine. 

L.cot 

D.l" 

L.  tang. 

54^ 

66 


TRIGOXOMETIilCAL    TUXCTIOXS. — 36°. 


Nat.  Functions. 

Logarithmic  Functions  +  10.              1 

0 

N.SlTlft 

N.  COS. 

L.  sine. 

D.  1" 

L.  COS. 

^! 

L.  tang. 

Dl." 

L.cot 

58779 

i  80902 

9.769219 

2 

§9 

9.907958 

.53 

g. 861261 

4-43 

10.138739 

60 

1 

58802  ;  80885 

769393 

2 

907866 

•  53' 

861527 

•43 

188473 

59 

2 

58826 1  80867 

769566 

2 

89 

907774 

.53 

861792 

42 

i382o3 

68 

3 

58849 
5S873 

8o85o 

769740 

2 

89 

907682 

•53: 

862038 

42 

13794a 

57 

4 

8o833 

769913 

2 

89 

907590 

.53 

862323 

42 

137677 

56 

5 

58896 

80816 

770087 

2 

1 

907498 

•53i 

862589 

42 

137411 

55 

6 

58920 

80799 

770260 

2 

907406 

.53: 

862854 

42 

137146 

54 

7 

58943 

80782 

770433 

2 

88 

907314 

-54 

863 1 19 

42 

i3688i 

53 

8 

58967 

80765 

770606 

2 

88 

907222 

•54 

863385 

42 

i365i5 

52 

9 

58990 

8s7.^8 

770779 

2 

88 

907129 

.54 

863650 

42 

i3635o 

51 

10 

59014 

80730 

770952 

2 

88 

907037 

-54 

863915 

42 

i36o85 

50 

-K 

11 

59037 j 80713 

9.771123 

2 

"SS' 

9.906945 

•54 

9^864180  1  4 

42 

10.135820 

49 

12 

59061^80696 

771298 

2 

87 

906852 

.54. 

864445 

4 

42 

135555 

43 

13 

59084  80679 

771470 

2 

87 

906769 

.54: 

864710 

4 

42 

135290 

47 

14 

59108,80662 

771643 

2 

87 

906667 

-54, 

864975 

4 

41 

135020 

46 

15 

59131  80644 

771815 

2 

87 

906575  1 

•54 

865240 

4 

41 

134760 

45 

16 

59154180627 

771987 

2 

87 

906482 

-54i 

8655o5 

4 

41 

134495 

44 

17 

59178 

80610 

772159 

2 

87 

906389  1 

.55, 

865770 

4 

41 

l3423o 

43 

X 

IS 

59201 

80593 

772331 

2 

86 

906296 

•55 

866035 

4 

41 

133965 

42 

19 

59225 

80576 

7725o3 

2 

86 

906204  1 

.55i 

866300 

4 

41 

133700 

41 

20 
21 

59248  8o558 

772675 

2 

86 

906111  1 

•  55: 

866564 

4 

41 

133436 

40 

59272  j8o54i 

9.772847 

2 

W 

9.906018  I 

"T55i 

9.866829 

4 

41 

10.133171 

3  J 

22 

59295 1 8o52< 

773018 

2 

86 

905925  1 

•  551 

867094 

4 

41 

132906 

33 

23 

59318 

8o5o7 

773190 

2 

86 

9o5S32  1 

•55 

867358 

4 

41 

132642 

37 

24 

59342 

80489 

773361 

2 

85 

900739  I 

.55 

867623 

4 

41 

132377 

36 

/ 

25 

5o365 

80472 

773533 

2 

85 

905645  I 

•  55! 

867887 

4 

41 

i32ii3 

35 

26 

59389 

80455 

773704 

2 

85 

905552  I 

•  55, 

868i52 

4 

40 

i3i848 

34 

27 

59412 

8o438 

773875 

2 

85 

905459  I 

•  55 

868416 

4 

40 

i3i584 

33 

23 

59436 

80420 

774046 

2- 

85 

9o5366  I 

•  56: 

868680 

4 

40 

l3i32o 

32 

29 

59459 

80403 

774217 

2- 

85 

905272  I 

•  56i 

868945 

4 

40 

i3io55 

31 

SO 

5?482 

8o386 

774388 

2 

84 

905 179  I 

•  56| 

869209 

4 

40 

130791 

30 

31  {  59506 

8o368 

9-774558 

2- 

84 

9.900085  I 

1^' 

9-869473 

4 

40 

io.i3o527 

'^y 

32  59529 

8o35i 

774729 

2- 

84 

904992  I 

•  56> 

869737 

4 

40 

i3o263 

23 

33 

59552 

80334 

774899 

2 

84 

904898  I 

-56; 

870001 

4 

40 

129909 

27 

34 

59576 

8o3i6 

775070 

2 

84 

904804  I 

.56 

870265 

4 

40 

129735 

26 

35 

59599 

80299 

775240 

2 

84 

9047 1 1  I 

.56 

870529 

4 

40 

129471 

25 

36 

59622 

80282 

775410 

2 

83 

904617  I 

.56 

870793 

4 

40 

129207 

21 

37 

59646 

80264 

775580 

2 

83 

904523  I 

.56 

871057 

4 

40 

128943 

23 

38  59669 

39  59693 

80247 

775750 

2- 

83 

904429  I 

•57 

871321 

4 

40 

128679 

22 

8o23o 

775920 

2 

83 

904335  I 

•57 

871585 

4 

40 

128415 

21 

40  1  59716 {80212 

776090 

2 

83 

904241  I 

•57; 

871849 

4 

39 

I28i5i 

20 

41  1  59739 1 80195 

9.776259 

2 

83 

9-904147  I 

-57' 

9.872112 

4 

39 

10.1278SS 

19 

42  '  59763  80178 

776429 

2 

82 

904053  I 

-57' 

872376 

4 

39 

127624 

13 

43  i  59786  80160 

776598 

2 

82 

903959  1 

•57: 

872640 

4 

39 

127360 

17 

44  ;  59809 '80143 

776768 

2 

82 

903864  • 

•57 

872903 

4 

39 

127097 

16 

45  t  59832  80125 

776937 

2 

82 

903770  I 

•57 

873T67 

4 

39 

126833 

15 

46  59856; 80108 

777106 

2 

82 

903676  I 

-57 

873430 

4 

39 

126570 

14 

47  1  59879 j 80091 

777275 

2 

81 

9o358i  1 

•57 

873694 

4 

39 

i263o6 

13 

4S  ,  59902 : 80073 

777444 

2 

81 

903487  I 

•57 

873957 

4 

39 

126043 

12 

49  1  59926 ; 8oo56 

777613 

2 

81 

903392  1 

.58 

874220 

4 

39 

123780 

11 

50 
51 

59949 ' 8oo38 

777781 

2 

81 

903 29S  I 

•58 

874484 

4 

39 

I255i6  1  10  1 

59972  80021 

9.777930 

2 

JT 

9-9o32o3  I 

•  08 

9-874747 

4' 

'39 

10. 125203 

9 

52  599q5 1 8ooo3 

7781 19 

2 

81 

903 1 o3  I 

•  58 

875010 

4 

39 

124990 

8 

53 

,60019! 799^^ 

778287 

2 

80 

9o3oi4  I 

•  58 

875273 

4 

38 

124727 

7 

54 

60042 ! 79968 

778455 

2 

80 

902919  I 

•  58 

875536 

4 

38 

124464 

6 

55 

6oo65  7995 1 

778624 

2 

80 

902824  1 

•  58 

875800 

4 

38 

124200 

5 

56  i  60089  79934 

778792 

2 

80 

902729  I 

•58 

876063 

4 

38 

123937 

4 

57  1  60112  i  79916 

778960 

2 

80 

992634  I 

•  58 

876326 

4 

38 

123674 

3 

53  60135:79899 
59  601 58  79881 

779128 

2 

80 

902539  I 

.59 

876589 

4 

38 

123411 

2 

779295 

2 

79 

902444  I 

.39 

876851 

4 

38 

I23i49 

1 

60  ^ 60182 j 79864 

779463 

2-79 

902349  • 

•  59, 

877"4 

4-38 

122886 

0 

N.  COS.  N.  sine. 

L.  COS. 

D.  1" 

L.  sine. 

' 

L.  cot.  1  D.  1" 

L.  tang.   '  1 

53^                         1 

trigono:metrical  functions. — 37^. 


e^l 


Nat.  Functions. 

Logarithmic  Functions  +  V).                                     j 

'  JN.sine. 

N.  COS. 

L.  sine. 

D.  1" 

L.  COS.   E 

M" 

L.  tang. 

D.  1" 

L.  cot. 

0  ''60182 

^867 

9-779463 

2-79 

9-902349  1 

i'> 

9-877114 

4.38 

10-122886 

60 

1  li  6o2o5 

79846 

779631 

2.79 

902253  I 

-59 

877377 

4-38 

122623 

oil 

2 

60228 

79829 

779798 

2-79 

9021 58  I 

•^9 

877640 

4-38 

122360 

58 

S 

6o25i 

79811 

779966 

2-79 

902063  I 

•59, 

877903 

4-38 

122097 

57 

4 

60274 

79793 

780133 

2.79 

901967  1 
901872  I 

19, 

878165 

4-38 

12i83d 

5*^ 

5 

60298 

79776 

780300 

2-78 

.59' 

878428 

4.38 

121572 

oo 

6 

,6o32i 

79758 

780467 

2.78 

901776  I 

.59 

87S691 

4-38 

i2i3o9 

54 

7 

6o344 

79741 

780634 

2.78 

901681  |i 

.59 

878953 

4-37 

121047 

53 

8 

60367 

79723 

780801 

2-78 

901685  |i 

19. 

879216 

4-37 

120784 

52 

9 

60390 

79706 

780968 

2.78 

901490  |i 

.59' 

879478 

4-37 

120322 

51 

10 

60414 

79688 

781134 

2.78 

901394  |i 

.60, 

879741 

4-37 

120259   .=^.(1  1 

11 

, 60437 

79671 

9.781301 

2-77 

9.90B298  ji 

.60 

9.880003 

4-37 

10-119997 

12 

' 60460 

79553 

781468 

2-77 

901202  I 

.60: 

880265 

4-37 

II9735 

13 

; 60483 

79635 

781634 

2.77 

901106  I 

.60 

88o528 

4-37 

119472 

14 

i  6o3o6 

79618 

781800 

2-77 

90101.0  ji 

.60' 

880790 

4-37 

II921O 

15 

i6o529 

79600 

781966 

2.77 

900914  I 

-60' 

88io52 

4-37 

16  :6o553 

79583 

782132 

2.77 

9008 1 8  I 

•60' 

88i3i4 

4-37 

1 1 8686 

44 

17  60676 

79565 

782298 

2.76 

900722  I 

•60 

881576 

4-37 

118424  4:3 

18 

: 60599 

79547 

782464 

2.76 

900626  I 

.60 

881839 

4-37 

118161   4-2 

19 

; 60622 

79530 

782630 

2.76 

000529  I 

•60 

882101 

4-37 

117899  41 
117637  40 

20  : 60645 

79512 

782796 

2.76 

900433  I 

•6ij 
~6\ 

882363 

4-36 

21  i  60668  7Q4Q4 

9-782961 

2-76 

9.900337  I 

9-882625 

4-36 

10-117375 

31" 

22  1  60691 

79477 

783127 

2.76 

900240  |i 

-61 

882887 
883148 

4-36 

II7113 

8S 

23 

60714 

79459 

783292 

2.75 

900144  [i 

•61 

4-36 

116852 

87 

24 

'60738 

79441 

783458 

2.75 

T°%  I 

-61 

883410 

4-36 

I 16590 

86 

25 

60761 

79424 

783623 

2.75 

.61, 

883672 

4-36 

II6328 

85 

26 

'  60784 

79406 

783788 

2.75 

899854  I 

.61: 

883934 

4-36 

116066 

34 

27 

60807 

79388 

783953 

2.75 

899757  I 

'^A 

884196 

4-36 

ii58o4 

S3 

28 

1 6o83o 

79371 

784118 

2.75 

899660  I 

-611 

884457 

4-36 

115543 

82 

29 

, 6o853 

79353 

784282 

2-74 

899564  I 

.61 

884719 

4-36 

ii528i 

81 

30 
31 

! 60876 

79335 

784447 

2-74 

899467  I 

.62 

884980 

4-36 

ll5020 

80 

60899 

79318 

9-784612 

2-74 

9-899370  I 

.62 

9-885242 

4-36 

10-114758 

21. 

32 

160922 

79300 

784776 

2-74 

899273  I 

.62 

8855o3 

4-36 

114497 

2S 

83 

60945 

79282 

784941 

2-74 

899176  I 

.62 

885765 

4-36 

114235 

27 

34 

60968 

79264 

785io5 

2-74 

.62 

886026 

4-36 

113974 

26 

85 

60991 

79247 

785269 

2.73 

.62| 

886288 

4-36 

113712 

25 

86 

■6ioi5 

79229 

785433 

2.73 

898884  I 

-62! 

886549 

4-35 

ii345i 

24 

87 

|6io38 

79211 

785597 

2.73 

898787  I 

.62i 

886810 

4-35 

1 13190  1  23 

88 

j6io6i 

79193 

785761 

2-73 

898689  I 

.62 

887072 

4-35 

112928  j  22 

89  1 61084 

79176 

785925 

2.73 

898592  I 

-62 

887333 

4-35 

112667  1  21 

40  '161107  I  79i58 

786089 

2.73 

898494  I 

-63 

887594 

4-35 

112406  '  20 

41  6ii3o 

79140 

9-786252 

2.72 

9-898397  I 

-63 

9-887855 

4.35 

10-112145  VJ 

42  J6ii53 

79122 

786416 

2.72 

898299  1 

•63) 

888116 

4.35 

111S84  '  IS 

43  61176 

79105 

786579 

2-72 

898202  I 

.63' 

^^■77 

4.35 

111623  17 

44  1161199 

79087 

786742 

2-72 

89S104  I 

.63 

888639 

4.35 

iii36i  1<) 

45  61222 

79069 

786906 

2.72 

898006  |i 

•  63 

888900 

4-35 

1 1 1 1 00   15 

46  ^61245 

7905 1 

787069 

2-72 

897908  I 
897810  I 

.63 

889160 

4.35 

110840 

14 

47  1 61268 

79033 

787232 

2.71 

.63, 

889421 

4.35 

110579 
iio3i8 

13 

48  ,61291 

79016 

787395 

2.71 

897712  I 

.63 

889682 

4-35 

12 

.49  ;6i3i4 

78998 

787557 

2.71 

897614  ji 

.63 

889943 

4-35 

110057 

11 

50  161337 

78980 

787720 

2-71 

897516  I 

-63 

890204 

4-34 

109-96 

10 

51  i!6i36o  178962 

9.787883 

2.71 

9.897418  I 

-64 

9.890465 

4.34 

10-109J35 

y 

52 

6i383 

78944 

788045 

2.71 

897320  I 

-64 

890725 

4-34 

109275 

s 

53 

61406 

78026 

788208 

2.71 

897222  I 

.64 

890986 

4-34 

109014 

10S753 

7 

54 

61429 

78908 

788370 

2.70 

897123 II 

.64 

891247 

4-34 

6 

55 

6i45i 

78891 

788532 

2-70 

897025  ;i 

.64 

891507 

4-34 

108493 

5 

56 

[61474 

78873 

788694 

2-70 

896926  ,1 

.64 

891768 

4-34 

108232 

4 

57 

161497 

78855 

788856 

2.70 

896828  I 

•64 

892028 

4-34 

107972 

3 

58 

6i52o 

78837 

789018 

2-70 

896729  I 

-64 

892289 

4-34 

107711 

2 

59 

6i543 

78819 

789180 

2-70 

896631  'i 

•64 

892549 

4-34 

107431 

1 

60 

61 566 

78801 

789342 

2.69 

896532  [i 

.64 

892810 

4-34 

107190 

0 

jN.  COS. 

N.  sine 

L.  COS. 

D.  1" 

L.  sine.  } 

L.  cot.   D.  1" 

L.  tang. 

' 

52° 

-       1 

68 


TEIGOXOMETKICAL   FUXCTIOXS. — 38°. 


Nat.  Functions. 

Logarithmic  Functions  +  10,              1 

0 

N.sine.!  N.  cos. 

L.  sine. 

D.l" 

L.  COS. 

D.l" 

L.  tang. 

D.l" 

Loot 

6i566i  78801 

9-789342 

2.69 

9-896532 

1-64 

9-892810 

4-34 

10-107190 

GO 

1 

61689 1 78783 

789504 

2.69 

896433 

1-65 

893070 

4-34 

106930 

59 

2 

61612 1 78765 

789665 

2.69 

896335 

1-65 

i   893331 

4-34 

106669 

5S 

3 

,61635178747 

789827 

2.69 

896236 

1-65 

893391 

4-34 

1 06409 

57 

4 

61658178729 

789988 

2-69 

896137 

1-65 

893831 

4-34 

106149 

56 

5 

161681  78711 

790 '49 

2-69 

896038 

1-65 

8941 1 1 

4-34 

105889 

55 

6 

61704  78694 

7903 10 

2-68 

895939 

1-65 

1   894371 

4-34 

105629 

54 

7 

61726178676 

790471 

2-68 

895840 

1-65 

1   894632 

4-33 

105368 

63 

8 

61749 1 78658 

790632 

2-68 

895741 

1-65 

1   894892 

4-33 

io5io8 

52 

9 

1.61772  78640 

790793 

2-68 

895641 

1.65 

895152 

4-33 

104848 

51 

10 

61795  78622 

790954 

2-68 

895542 

1-65 

1-66 

I   895412 

<-33 

104588 

50 

11 

61818  78604 

9.791115 

2-68 

9-895443 

:  9-895672 

4-33 

10.104328 

49 

12 

6 I 84 I  178586 

7912-5 

2-67 

895343 

1-66 

895932 

4-33 

1 04068 

48 

13 

61864  S  78568 

791436 

2-67 

893244 

1-66 

896192 

4-33 

io38o3 

47 

14 

61887178550 

791596 

2-67 

895145 

1-66 

896452 

4-33 

103548 

46 

15 

61909  78532 

791757 

2-67 

895045 

1-66 

896712 

4-33 

103288 

45 

16 

61932 

78514 

791917 

2.67 

894945 

1-66 

895971 

4-33 

io3o29 

44 

17 

61933 

78496 

792077 

2.67 

894846 

1-66 

897231 

4-33 

102769 

43 

IS 

61978 

78478 

792237 

2-66 

894746 

1-66 

897491 

4-33 

102509 

42 

I'J 

62001 

78460 

792397 

2-66 

894646 

1-66 

897751 

4-33 

102249 

41 

20 

62024 

78442 

792557 

2-66 

894546 

1-66 

898010 

4-33 

101990 

40 

::l 

62046" 

78424 

9-792716 

2-66 

9-894446 

1.67 

1  9-89S270 

4-33 

10-  ion3o 

39 

li:: 

62069 

78405 

792876 

2-66 

894346 

1.67 

898530 

4-33 

101470 

3S 

23 

62092 

78387 

793o35 

2-66 

894246 

1.67 

89S789 

4-33 

101211 

37 

24 

62115 

78369 

793195 

2-65 

894146 

1-67 

II& 

4-32 

100951 

36 

25 

62138 1 78351 1 

793354 

2-65 

894046 

1-67 

4-32 

100692 
100432 

35 

26 

62160 

78333 

7935i4 

2-65 

893946 

1-67 

899568 

4-32 

34 

27 

62183 

7831 5 

793673 

2-65 

893846 

1-67 

899827 

4-32 

100173 

33 

2S 

62206 

78297 

793832 

2-65 

893745 

1-67 

900086 

4-32 

099914 

32 

2& 

62229 

78279 

793991 

2-65 

893645 

1-67 

900346 

4-32 

099654 

31 

oO 

6225l 

78261 

794 1 5o 

2-64 

893544 

1-67 

900605 

4.32 

099395 

30 

ol 

62274178243 

9 -794308 

2-64 

9-893444 

1-68 

9-900864 

4-32 

10-099136 

29 

32 

62297  78225 

794467 

2-64 

893343 

1-68 

901124 

4-32 

098876 

23 

33 

62320 1 78206 

794626 

2-64 

893243 

1-68 

901383 

4-32 

098617 

27 

34 

62342  1 78188 

794784 

2-64 

893142 

1-68 

901642 

4-32 

098358 

26 

35 

62365  '78170 

794942 

2-64 

893041 

1-68 

901901 

4-32 

098099 

25 

36 

62388  78152 

795101 

2-64 

892940 

1-68 

902160 

4-32 

097840 

24 

37 

62411  78134 

795259 

2-63  • 

892839 

1-68 

902419 

4-32 

097581 

23 

3S 

62433  781 16 

795417 

2-63 

892739 

1-68 

902679 
902938 

4-32 

097321 

22 

39 

62456  i  78098 

795575 

2-63 

892638 

1-68 

4.32 

097062 

21 

40 

62479  78079 

795733 

2-63 

892536 

1-68 

903197 

4-3i 

096803 

20 

41 

62502  178061 

9-795891 

2-63 

9-892435 

1-69; 

9 -903455 • 

4-3i 

10-096545 

19 

42 

62524  i 78043 

796049 

2-63 

892334 

1-69 

903714 

4.31 

096286 

IS 

43 

62547  '<  78025 

796206 

2-63 

892233 

1-69 

903973 

4-3i 

096027 

17 

44 

62570  1  78007 

796364 

2-62 

892132 

1-69 

904232 

4-3i 

095768 

16 

45 

62592  J7798S 

796321 

2-62 

892030 

1-69 

904491 

4-31 

095509 

15 

46 

62615  , 77970 

796679 

2-62 

891929 

1-69I 

904730 

4-31 

095250 

14 

47 

62638  77952 

796836 

2-62 

891827 

1.691 
1-69 

903008 

4-31 

094992 

13 

48 

62660  1 77934 

796993 

2-62 

891726 

905267 

4-3i 

094733 

12 

49 

62683  1  77916 
62706  i  77897 

797 i5o 

2-61 

891624 

1-69! 

905526 

4-3i 

094474 

11 

50 

797307 

2-6l 

891523 

1-70, 

905784 

4-3i 

094216 

10 

IT" 

62728  77879 

9-797464 

2-61 

9-891421 

I -70! 

9.906043  4-3i  1 

10-093937 

9 

52 

62731 

77861 

797621 

2-61 

891319 

1-70, 

9o63o2 

4-3i 

093698 

8 

53 

62774 

77843 

797777 

2-61 

891217 

1.70 

906360 

4-3i 

093440 

7 

54 

62796 

77824 

797934 

2.61 

891115 

1-70, 

906819 

4-3i 

093181 

6 

5d 

62819 

77806 

798091 

2-61 

891013 

i-7o| 

907077 

4-31 

092923 

5 

56 

62842 

77788 

798247 

2-61 

890911 

1-70 

907336 

4.31 

092664 

4 

57 

62864  1  77769 

798403 

2-60 

890809 

1.70 

907594 

4-31 

092406 

3 

58 

62887  77751 

798560 

2 -60 

890707 

1-70, 

907852 

4.31 

092148 

2 

59 

62909 

77733 

798716 

2.60 

890605 

1-70 

908111 

4-3o 

091889 

1 

60 

62932 

77715 

798872 

2-60 

89o5o3 

1.70 

908369 

4.30 

091631 

0 

N.  co8.,N.8ineJ 

L.  COS. 

D.l" 

L.  sine. 

! 

L.  cot. 

D.l" 

L.  tang. 

' 

51°                         1 

TRIGONOMETRICAL   F  CNCTIONS. — 39=" 


69 


Nat.  Functions. 

Logarithmic  Functions  +  10.              1 

'   N.sine.|N.  COS. 

L.  sine. 

D.  1" 

L.  COS.   I 

).l" 

L.  tang. 

Dl." 

L.  cot. 

60 

■ 
0 

62932 

77715 

9.798872 

2-60 

9-890503  I 

.70 

9-908369 

4-3o 

10-091631 

1 

62955 

77696 

799028 

2 -60 

890400  1 

•71 

908628 

4 

3o 

091372 

59 

2 

62977 

77678 

799184 

2 -60 

890298  I 

•7' 

908886 

4 

3o 

091 1 14 

58 

3 

63ooo 

77660 

799339 

2.59 

890195  I 

•71 

909144 

4 

3o 

.090856 

57 

4 

63022 

77641 

799495 

2.59 

890093  I 

•7' 

909402 

4 

3o 

090598 

56 

5 

63045 

77623 

799631 

2-59 

'&i ; 

•71 

909660 

4 

3o 

090340 

55 

6 

63o68 

77605 

799806 

2.59 

•7' 

909918 

4 

3o 

090082 

54 

7 

63090 

77586 

799962 

2.59 

889785  I 

•71 

910177 

4 

3o 

089823 

53 

8 

63ii3 

77568 

800117 

2.59 

889682  I 

•71 

910435 

4 

3o 

089365 

52 

9 

63i35 

77550 

800272 

2-58 

889579  1 

•7" 

9106^3 
910931 

4 

3o 

089307 

51 

10 

63 1 58 

7753i_ 

800427 

2-58 

889477  ' 

•71 

4 

3o 

089049 

50 

4y 

11 

63 1 80 

775i3 

9.800582 

2-58 

9-889374  I 

••72 

9-911209 

4 

3o 

10-088791 

12 

63203 

77494 

800737 

2.58 

889271  I 

•72 

911467 

4 

3o 

088533 

48 

13 

63225 

77476 

800892 

2.58 

889168  I 

•72 

911724 

4 

3o 

088276. 

47 

14  1 

632^8 

77458 

801047 

2-58 

889064  I 

•72 

911982 

4 

3o 

088018 

46 

15  i 

63271 

77439 

801201 

2.58 

888961  I 

•72 

912240 

4 

3o 

087760 

45 

16  1 

63293 

77421 

8oi356 

2-57 

888858  1 

•72 

912498 

4 

3o 

087502 

44 

17  1 

633 16 

77402 

8oi5ii 

2.57 

888755  I 

.72 

912756 

4 

3o 

087244 

43 

18 

63338 

77384 

8oi665 

2.57 

888651  1 

.72 

9i3oi4 

4 

29 

086986 

42 

19 

63361 

77366 

801819 
801973 

2.57 

888548  1 

.72 

913271 

4 

29 

086729 

41 

'20 

63383 

77347 

2.57 

888444  I 

•73 
•73 

913529 

4 

29 

086471 

40 
"3'7 

21 

63406 

77329 

9.802128 

2.57 

9- 88834 1  I 

9-913787 

4 

29 

io-o862i3 

22 

63428 

77310 

802282 

2-56 

888237  « 

•73 

914044 

4 

29 

085956 

33 

23 

63451 

77292 

802436 

2.56 

888134  I 

•73 

914302 

4 

29 

085698 

37 

24 

63473 

77273 

802089 

2-56 

888o3o  I 

•73 

914560 

4 

29 

085440 

36 

25 

634Q6 

77255 

802743 

2-56 

887926  I 

•73 

914817 

4 

29 

b85i83 

35 

26 

635i8 

77236 

802897 

2-56 

887822  I 

•73 

915075 

4 

29 

084925 

34 

27 

63540 

77218 

8o3o3o 

2-56 

887718  1 

.73 

913332 

4 

29 

084068 

33 

28 

63563 

77199 

8o32o4 

2-56 

887614  I 

•73 

915390 

4 

29 

084410 

32 

29 

63585 

77181 

803357 

2.55 

887510  I 

•73 

•   915847 

4 

29 

084153 

31 

30 
31 

636o8 

77162 

8o35ii 

2.55 

887406  1 

•74 

916104 

4 

29 

083896 

80 
29 

63630 1 77144 

9.803664 

2-55 

9-887302 

9-916362 

4 

29 

10-083638 

32 

63653 

77125 

8o38i7 

2.55 

887198  1 

•74 

916619 

4 

29 

o8338i 

28 

33 

63675 

77'07 

803970 

2-55 

887093  1 

•74 

916877 

4 

29 

083.23 

27 

34 

63698 

770S8 

804123 

2-55 

886989 

•74 

917134 

4 

29 

082866 

26 

35 

63720 

77070 

804276 

2.54 

886885  I 

•74 

917391 

4 

29 

082609 

25 

36 

63742 

77o5i 

804428 

2.54 

886780  1 

•74 

917648 

4 

29 

082352 

24 

37 

63765 

77033 

804581 

2-54 

886676 

•74 

917905 

4 

29 

082095 

23 

38 

63787 

77014 

804734 

2.54 

886571 

•74 

918163 

4 

28 

081837 

22 

39 

63810 

76996 

804886 

2.54 

886466 

•74 

1   918420 

4 

28 

o3i58o 

21 

40 
41 

63832 

76977^ 

8o5o39 

2.54 

886362 

•75 
•75 

9 '8677 

4 

28 

o8i323 

20 

63854 

76939 

9-8o5i9i 

2-54 

9-856257 

!  9-918934 

4 

28 

10-081066 

19 

42 

63877 

76940 

805343 

2-53 

836i52 

•75 

919191 

4 

28 

080809 

18 

43 

,63899 

76921 

805495 

2-53 

886047 

•■'^ 

919448 

4 

28 

o8o552 

17 

44 

63922 

76903 

8o5647 

2.53 

883942 
885837 

•75 

919705 

4 

28 

080295 
o8oo38 

16 

45 

63944 

76884 

til?: 

2.53 

•75 

919962 

4 

28 

15 

46 

63966 

76866 

2.53 

885732 

■■'^ 

920219 

4 

28 

079781 

14 

47 

•  63989 

76847 

806 I o3 

2-53 

885627 

•75 

920476 

4 

28 

079524 

13 

48 

(64011 

76828 

806254 

2.53 

885522 

•75 

920733 

4 

28 

079267 

12 

49 

i  64033 

76S10 

806406 

2.32 

885416 

•75 

920990 

4 

28 

079010 

11 

r50 

! 64056 

76791 

806557 

2-52 

8853 1 1 

-76 

1   921247 

4 

28 

078753 

10 

51 

1 64078 

76772 

9-806700 

~J^5T 

9-885205 

-76 

1  9 -921503 

4 

28 

10-078497 

9 

52 

164100 

76754 

806860 

2-52 

885ioo 

-76 

921760 

4 

28 

078240 

8 

53 

64123 

76735 

807011 

2-52 

884994 

•76 

922017 

4 

28 

077983 

7 

54 

64145 

76717 

807163 

2.52 

884889 
884783 

•76 

922274 

4 

28 

077726 

0 

55 

64167 

76698 

807314 

2-52 

1-76 

922530 

4 

28 

077470 

5 

56 

64190 

76679 

807465 

2-51 

884677 

-76 

922787 

4 

28 

077213 

4 

57 

64212 

76661 

807615 

2.5l 

884572 

-76 

923044 

4 

28 

076956 

8 

58 

64234 

76642 

807766 

2-51 

884466 

-76 

923300 

4 

28 

076700 

2 

59 

64256 

i  76623 

807917 

2-51 

884360 

-76 

923557 

4 

27 

076443 

1 

60 

64279  76604 

808067 

2.5l 

884254 

'•77 

9238i3 

4-27 

076187 

0 

■ 

|N.  COS.  N.  sine 

L.  COS. 

D.  1" 

L.  sine. 

!|  L.cot. 

D.l" 

L.  tang. 

50°                         1 

70 


TRIGONOMETRICAL   FUXCTIOXS. — 40°. 


Nat.  FrxcTioNs. 

LOGAKITHailC  FlTNCTIOKS  +  10. 

'  iN.sme.'N.  cos.|  L.  sine. 

D.  1" 

L.  COS.  JD.l' 

L.  tang. 

D.l" 

1   L.  cot 

0l|  64279 '76604 

9.808067 
808218 

2-5l 

9-884254  11.77 

9.923813 

4-27 

110.076187 

60 

1  1  64301  I  76586 

2-5l 

884148  ;i.77 

924070 

4-27 

1   075930 

59 

2  !  64323 1 76367 

8o8368 

2-5l 

884042  1-77 

924327 

4-27 

075673 

53 

s  !  64346 1 76548 

8o85i9 

2-5o 

883936  '1.77 

924583 

4-27 

075417 

57 

4  ;  64368.76530 

808669 

2-5o 

883829  1-77 
883723  1-7? 

924840 

4-27 

075160 

56 

5  64390^76511 

808819 

2-5o 

925096 

4.27 

074904 

55 

6  ; 64412 ,76492 

808969 

2-5o 

883617  ..77 

925352 

4-27 

07464S 

54 

7  64433 

76473 

8091 19 

2-5o 

883510  1-77 

925609 
925865 

4-27 

074391 
074135 

53 

8  64457 

76455 

809269 

2-5o 

883404  ;i-77 

4-27 

52 

9  64479176436 

809419 

2-49 

883297  :i-78 

926122 

4-27 

073878 

51 

10  1  64301 ,76417 

809569 

2-49 

883I9I  J -78 

926378 

4-27 

073622 

50 

11  1  64524  i  76398 

12  j  64546  7638o 

9-809718 

2-49 

1  9-883o84  I .78:  9-926634 

4-27 

10.073366 

49 

809868 

2-49 

882977  1-78 
882871  1.78 

926890 

4.27 

073110 

48 

13  i  645681 76361 

810017 

2-49 

927147 

4-27 

072853 

47 

14  i  64590  <  76342 

810167 

2-49 

882764  !i.78 

927403 

4.27 

072597 

46 

15  ,64612:76323 

8io3i6 

2-48 

882657  ;i-78 

927659 
927913 

4.27 

072341 

45 

16  ,64635  176304 

810465 

2-43 

882550  '1-78 

4-27 

072085 

44 

17  i:  64657 

76286 

810614 

2-43 

S82443  ,1-78 

928171 

4-27 

071829 

43 

18  ;  64679 

76267 

810763 

2-43 

8ftj336  1-79 

928427 

4-27 

071573 

42 

19  1  64701 

76248 

810912 

2-43 

882229  ,^'79 

028683 

4-27 

071317 

41 

20  64723 

76229 

811061 

2-43 

8S2121  1-79!   928940 

4-27 

071060 

40 

oi '  i'^n^o 

96210 

9-811210  2-43 

9-882014  1-79  9.929196 
881907  ,1.79;   929452 

4-27 

10-070804 

89 

11 !  Af  ^^ 

76192 

8ii358  2-47 

4-27 

070548 

88 

23  1  64790 

76173 

8n5o7  2-47 

881799  1.79!   Q29708 

4-27 

070292 
070036 

87 

24  i  64812 

76154 

8ii655  2-47 

881692  1-79 

929964 

4-26 

36 

25  648341 76135 

81 1804 

2-47 

88i584  1.79 

930220 

4-26 

069780 

85 

26  j  64856 '761 16 

811932 

2-47 

881477  ,1-79 

930475 

4.26 

069525 

84 

27  i  64878 

76097 

812100 

2-47 

88,369  1.79 

930731 

4-26 

069269 

33 

28  i  64901 

76078 

812248 

2-47 

881 261  1-80 

930987 

4-26 

069013 

82 

29  64923 

76059 

812396 

2-46 

88ii53  1-80 

931243  4-26 

068757  81 

30  64945  76041  1 

812544 

2-46 

881046  ;i-8o 

931499  4-26 

o685oi  SO 

81  ,,64967 

76022 

9-812692 

2-46 

9.88093s  1.80 

9.931755 

4-26 

10-068245  29 

32  ||  64989 

76003 

812840 

2-46 

88o83o  '1-80 

932010 

4.26 

067990 

28 

33  1;  65oi  I 

75984 

812988 

2-46 

880722  1-80 

932266 

4-26 

067734 

27 

34  :  65o33 

75963 

8i3i35 

2-46 

880613  1-80 

932522 

4.26 

067478 

26 

35  65o55 1 75946 

8i3283 

2-46 

88o5o5  1-80 

932778 

4-26 

067222 

25 

36  ,63077175927 

8i343o 

2-43 

880397  |i-8o 

933o33 

4-26 

066967 

24 

37  1  63100  73908 

813578 

2-45 

880289  i-8i 

933289 
933543 

4.26 

066711 

23 

33  ;  65i22| 75889 

813725 

2-45 

880180  1-81 

4-26 

066455 

22 

39  65i44l 753-0 

813872 

2-43 

880072  1-81 

933800  4-26 

066200 

21 

40  65i66.75S5i 

814019 

2-45 

879963  1-81 

934056  4-26 

063944 

20 

41  65i88 175832 

9-814166 

2-45 

9-879855  I -81 

9.934311 

4-26 

10-0656S9 
065433 

19 

42  65210 |758i3 

8i43i3 

2-43 

879746  :i-8r 

934567 

4.26 

18 

43  ;  65232  75794 

814460 

2-44 

879637  1-81 

934823 

4.26 

063177 

17 

44  1!  65254   '  " 

73773 

814607 

2-44 

879529  I -81 

935078 

426 

064922 

16 

45  1  65276 

75756 

814753 

2-44 

879420  I -81 

935333 

4.26 

064667 

15 

46  65298 

73738 

814900 

2-4< 

879311  I -81 

935589 

4.26 

0644 II 

14 

47  1  65320 j 

75719 

816046 

2-44 

879202  1-82 

935844 

4.26 

0641 56 

13 

43  ,  65342 1 73700 

8i5i93 

2-44 

879093  1.82 

936100 

4-26 

063900 

12 

49  65364  j  7568o 

815339 

2-44 

878984  1-82 
878875  '1.82 

936355 

4.26 

063645 

11 

50  ,65386  175661 

81 5485 

2-43 

936610 

4-26 

063390 

10 

51  ,65408175642 

9-8i563i 

2-43 

9-878766  J -82; 

9-936866 

4-25 

io-o63i34 

9 

52  6543o 1 75623 

815778 

2-43  : 

878656  1.82  1 

937121 

4-2^ 

4-23 

062879 

8 

53  65452 ! 75604 

815924 

2-43 

878547  1-82 

937376 

062624 

7 

54  '65474175585 

816069 

2-43 

878438  1-82 

937632 

4-25 

062368 

6 

55  163496 

75566 

8i63i5 

2-43 

878328  1-82 

937887 

4.25 

0621 i3 

5 

56  '  655i8 

75547 

8i636i 

2-43 

878219  i-83| 

938142  4-25 

06 I 858 

4 

57  65540 

75528 

8i65o7 

2-42 

878109  1-83, 

938398  1  4.25 

061602 

8 

58  65562 1 75509 

8i6652 

2-42 

877999  I -831 

938653  4-25 

061347 

■2 

59  65584 ■  75490 

816798 

2-42 

877890  I -83! 
877780  I -831 

938908  4-25 

061092 

1 

60  656o6l 75471 

816943 

2.42 

939163  1  4-25 

060837 

0 

N.  COS.  N.  sine. 

L.  COS.   B.  1"  ' 

L.  sine.  1    '1  L.  cot.   D.  1"  | 

L.  tang. 

49°                          1 

TRIGONOMETRICAL    FUXCTIOXS. — 41' 


71 


Na^.  Functions. 

Logarithmic  Functions  +  10.              1 

' 

N.sine.,  N.  cos 

L.  sine. 

D.l" 

L.  COS. 

D.l" 

1  L.  tang. 

D.l" 

1  L.  cot 

0 

l656o6| 75471 

9.816943 

2.42 

9.877780 

1.83 

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4-25 

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3 

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4 

' 65694 

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1-83 

940183 

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5 

65716 

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1-84 

940438 

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6 

; 65738 

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65759 

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53 

8 

1 65781 

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9 

! 658o3 

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10 

165825 

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2-41 

876678 

1-84 

941714 

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058286 

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ir 

65847 

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9-8i8536 

2-40 

9.876568 

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■   9.941968 

4-25 

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12 

165869 

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876457 

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942223 

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13 

J  65891 

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1  9-944517 

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22 

66088 

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944771 

4-24 

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23 

66109 

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945026 

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1.86 

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66 1 53 

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1-86 

945535 

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2-36 

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44 

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17 

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49 

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1.88 

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2.35 

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1-88 

951896 

4-24 

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10 

51 

66718 

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9-824245 

2-35 

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1-89 

9.952150 

4-24 

10-047850 

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52 

66740 

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824386 

2.35, 

1.89 

952405 

4-24 

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53  , 

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824527 
824668 

2.35 

871868 

1-89! 

952659 

4-24 

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7 

54 

iii^i 

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2-34 

871755 

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952913 

4-24 

047087 

6 

55 

66805 

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2-34 

871641 

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953167 

4-23 

046833 

5 

56 

66827 
66848 

74392 

824949 

2-34 

871528 

1.89 

953421   4-23 

046579 
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4 

57 

74373 

825090 
825230 

2-34 

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1.89 

953675  4-23 

8 

58 

66870 

74353 

2-34 

871301 

1.89 

953929  4-23 

046071 

2 

59 

66891 

74334 

825371 

2-34 

871187 

1-89 

954183  4-23 

045817 

1 

eo 

66913 

74314 

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2-34 

871073 

1.90 

954437 

4-23 

045563 

0 

N.  COS.  N.  sine.' 

L.  COS.   D.  1" 

L.  sine. 

L.  cot. 

D.l" 

L.  tang. 

f 

48° 

'1 

72 


TEIGOXOMETRICAL    FUXCTIOXS. — 42^^ 


Nat,  FxrNCTioNs. 

LOGAKITHMIC  FUNCTIOXS  +  10 

1 

0 

N.  sine.!  N.  COS. 

L.  sine.  |  D.  1" 

L.COS. 

'r 

1  L.tang. 

D 

.  1" 

Kcot  1 

66913I74314 

9-825511   2 

•34 

9-871073 

1.90 

;  9-954437 

7 

23 

IO-045563  ,  60 

1 

66935  74295 

825651  ]  2 

-33 

870960 
870846 

1-90 

1  954691 

4 

23 

045309  59 

2 

66956 j 74276 

825-91  1  2 

-33 

1.90 

954945 

4 

23 

o45o53  ■   5^. 

3 

66978 

74256 

825931  ,  2 

•33 

870732 

1-90 

955200 

4 

23 

044800  57 

4 

66999 

74237 

826071  1  2 

.33 

870618 

1.90 

955454  j  4 

23 

044546 

56 

£ 

67021 

74217 

826211 

2 

33 

870504 

1.90 

955707  4 

23 

044293 
044039 

55 

6 

67043 

74198 

826351 

2 

33 

870390 

1-90 

955961  4 

23 

54 

7 

blobi 

74178 

826491  i  2 

33 

870276 

1.90 

956213  !  4 

23 

043783 

53 

8 

67086 

74159 

826631  !  2 

33 

870161 

1-90 

956469 
956723 

4 

23 

04353 1 

52 

9 

67107 

74139 

826770  j  2 

32 

870047 

^•9. 

4 

23 

043277 

51 

10 

67129 

74120 

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32 

869933 

1^1 

936977 

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23 

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11 

67151 

74100 

9-82-049  2 

32 

9-869818 

1-91 

9-957231 

4 

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10-042769 

4y 

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67172 '74080 

82-189  2 

32 

869704 

1-91 

957485 

4 

23 

o425i5 

43 

13 

67194  74061 

82-328  I  2 

32 

869589 

1-91 

957739 

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23 

042261 

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67215; 74041 

827467 

2 

32 

86Q474 

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957993 

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23 

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827606 

2 

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1-91 

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23 

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16 

67258 

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32 

869245 

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17 

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1-91 

958754 

4 

23 

041246 

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18 

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1-92 

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23 

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1-92 

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4 

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41 

20 

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2 

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1-92, 

959516 

4 

23 

040484 

40 

21 

67306  73904 

9.828439 
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2 

3i 

9-868670 

1-92; 

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23 

10-040231 

39 

22 

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2 

3i 

868555 

1-92; 

4 

23 

039977 

£3 

2S  I 6T4CQ 173863 

828716 

2 

3i 

868440 

1-92 

960277 

4 

23 

039723 

37 

24 

167430  73846 

828855 

2 

3o 

868324 

1-92 

00531 

4 

23 

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039216 

36 

25 

1 67452  -73826 

828993 

2- 

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868209 

1-92: 

960784 

4 

23 

35 

26 

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2 

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1-92 

961038 

4 

23 

038962 

34 

27 

67495173787 

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2 

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1-93 

961291 

4 

23 

038709  1  33 

28 

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23 

038453  1  32 

29  '67538  73747 

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31 

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829959  1  2 

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867309 
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1.93 

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29 

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4 

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34  67645. 7364q[ 

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1.93 

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2 

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4 

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37 

' 67709 1 73590 

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1-94 

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4 

23 

036173  1  23 

83 

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i^§4 

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4 

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39  67702  73551 

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40  :  67773  73531 1 

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1:941 

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22 

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20 

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9-831195 

2 

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9-866353 

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9-964842  I  4 

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io-o35i58 

19 

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2 

23 

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1.94: 

960095  i  4 

22 

o349o5 

18 

43 

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2- 

28 

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1.94: 

965349  i  4 

22 

o3465i 

17 

44 

67859 '73452 

83 1606 

2- 

28 

866004 

1-95; 

965602 

4 

22 

034398 

16 

4? 

67880,73432 

831742 

2- 

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865887 

1.95 

965855 

4- 

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15 

4f 

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83.1879 
832015 

2- 

28 

865770 

1.95; 

966109 

4 

22 

033891 
033638 

14 

47 

67923  73393 

2 

27 

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1 .95, 

966362 

4- 

22 

13- 

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2- 

27 

865536 

966616 

4- 

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033384 

12 

49 

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2- 

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11 

50 

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4- 

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10 

51 

68008 '73314 

9-832561  ;  2 

27 

9 -865 1 85 

1.95 

9-967376 

4- 

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10-032624 

9' 

52 

6S029  i  73294 

832697 

2 

27 

86D068 

i-95i 

Zslt 

4- 

22 

032371 

8 

53 

68g5i i  73274 

832833 

2- 

27 

864950 

1-95; 

4- 

22 

032117 

7 

54 

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832969 

2 

26 

864833 

1  -961 

968136 

4 

22 

o3i864 

6 

55 

68093:73234 

8331 03 

2 

26 

864716 

1-961 

968389 

4 

22 

o3i6ii 

5 

56 

'68ii5i732i5 

833241 

2 

26 

864598 
864481 

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968643 

4 

22 

o3i357 

4 

57  ,68i36 173195 

833377 

2 

26 

1-96; 

968896 

4 

22 

o3iio4 

8 

53  .i68i57  73175 

.  833512  2 

26 

864363 

1-96; 

969149 
969403 

4 

22 

o3o85i 

2 

59 

,68179  73 1 55 

833648  2 

26 

864245 

..96 

4 

22 

o3o597 

1 

60 

68200,73135 

833783  2-26 

864127 

1.96! 

969656 

4-22 

o3o344 

0 

N.  COS.  N.sine. 

L.  COS.   '  D.  1" 

L.  sine. 

^ 

L.  cot   1  D.  1" 

L.  tang. 

~^ 

1 

TRIGONOMETRICAL   FUXCTIOXS. — 43^ 


73 


Nat.  Functions. 

LOGAKITHMIC  FUNCTIONS  +  10.                    1 

' 

N.sine.i  N.  cos. 

1 

L.  sine.  1  D.  1" 

L.  COS.  |l 

).i" 

'  L.  tang. 

D.  1" 

L-  cot. 

0 

68200 

73,35 

9-833783 

2 

26 

9-864127  I 

.96 

:  9-969656 

4-22 

io-o3o344 

60 

1 

68221 

73116 

8339,9 

2 

25 

864010  I 

-96 

1   969909 

4-22 

030091 

59 

2 

6S242 

73096 

834054 

2 

25 

863892  I 

•97 

1   970162 

4-22 

029838 

58 

8 

, 68264 

73076 

834189 

2 

25 

863774  1 

•97 

1   970416 

4-22 

029584 

57 

4 

i 68285 

73o56 

834325 

2 

25 

863656  J 

■97 

970669 

4-22 

029331 

56 

5 

i  683o6 

73o36 

834460 

2 

25 

863538  1 

•97 

970922 

4-22 

029078 

55 

6 

] 68327 

73016 

834595 

2 

25 

863419 

•97 

971175 

4-22 

028825 

54 

7 

68349 

72996 

834730 

2 

25 

863301 

•97 

971429 

4-22 

028571 

53 

8 

68370 

72976 

834865 

2 

25 

863 1 83 

•97 

97,682 

4-22 

0283,8 

52 

9 

6839. 

72957 

834999 

2 

•24 

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•97 

97,935 

4-22 

028065 

51 

10 

68412 

72937 

833,34 

2 

24 

862946 

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-98 

972,88 

4-22 

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50 

11 

68434 

72917 

9-835269 

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24 

9-862827 

1  9^972441 

4-22 

10-027559  1  49  1 

12 

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72897 

835403 

1 

24 

862709 

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972694 

4-22 

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4S 

13 

68476 

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2 

24 

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972948 

4-22 

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47 

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68497 

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2 

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4-22 

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15 

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4-22 

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16 

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9-3707 

4-22 

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17 

68561 

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2 

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4-22 

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18 

68582 

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4-22 

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2<;» 

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2 

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9-974973 

4-22 

10-02D027 

39 

22 

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2 

23 

86,5,9 

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975226 

4-22 

024774 

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23 

68688 

72677 

836878 

2 

23 

86,400  , 

•99 

975479 

4-22 

024521 

37 

24 

68709 

72657 

8370,2 

2 

22 

86,280 

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975732 

4-22 

024268 

36 

25 

68730 

72637 

837,46 

2 

22 

86,161  , 

•99 

975985 

4-22 

0240,5 

35 

26 

68751 

726,7 

837279 

2 

22 

861041  1 

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976238 

4-22 

023762 

34 

27 

68772 

72597 

837412 

2 

22 

860922  , 
860802  I 

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976491 

4-22 

023509 

33 

28 

68793 

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2 

22 

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976744 

4-22 

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2 

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4-22 

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30 
31 

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2 

22 

86o562  2 

-00 
-00 

977250 

4-22 

022750 

80 

68857 

72517 

9-837945 

2 

22 

9-860442  2 

9 -977503 

4-22 

10-022497 

29 

32 

68878 

72497 

838078 

2 

2, 

86o322  2 

-00 

977756 

4-22 

022244 

28 

38 

68899 

72477 

8382,1 

2 

21 

860202  2 

•00 

978009 

4-22 

02,99, 
02,738 

27 

34 

68920 

72457 

838344 

2 

2, 

860082  2 

•00 

978262 

4-22 

26 

35 

68941 

72437 

838477 

2 

2, 

859962  2 
859842  2 

-00 

9785,5 

4-22 

02,485 

25 

86 

68962 

72417 

8386,0 

2 

2, 

•  00 

978768 

4-22 

02,232 

24 

37 

68983 

72397 

838742 

2 

2, 

859721  2 

-01 

979021 

4-22 

020979 

23 

38 

69004 

72377 

838875 

2 

2, 

839601  2 

-01 

979274 

4-22 

020726  !  22  1 

39 

69025 

72357 

839007 

2 

2, 

859480  2 

•0, 

979527 

4-22 

020473 

21 

40 

69046 

72337 

839140 

2 

20 

859360  2 

-0, 

979780 

4-22 

020220 

20 
19 

41 

69067 

72317 

9-839272 

2 

20 

9-859239  2 

-01 

9-980033 

4-22 

10-0,9967 

42 

69088 

72297 

839404 

2 

20 

859,19  2 

-01 

980286 

4-22 

0197,4 

18 

43 

69109 

72277 

839536 

2 

20 

858998  2 

-0, 

980538 

4-22 

0,9462  17  1 

44 

6qi3o 

72257 

839668 

2 

20 

858877  2 

-0, 

980791 

4-21 

0,9209 

16 

45 

69151 

72236 

839800 

2 

20 

858756  2 

-02 

98,044 

4-21 

0,8956 

15 

46 

69172 

72216 

.  839932 

2 

20 

858635  2 

-02 

981297 

4-21 

0,8703 

14 

47 

69193 

72,96 

840064 

2 

'9 

8585,4  2 

•02 

98,550 

4-21 

o,845o 

18 

48 

69214 

72176 

840196 

2 

'9 

858393  2 

•02 

98,803 

4-2, 

0,8,97 

12 

49 

69235 

72156 

840328 

2 

'9 

858272  2 

-02 

982056 

4-21 

017944 

11 

5<> 

69256 

72,36 

840459 

2 

19 

858i5i  2 

-02 

982309 

4-21 

017691 

10 

51 

69277 

721,6 

9-840591 

2 

19 

9-858029  2 

•02 

Q-982562 

4-21 

10-017438 

9 

52 

69298 

72095 

840722 

2 

19 

857908  2 

-02 

9828,4 

4-21 

0,7,86 

8 

53 

69319 

72075 

840854 

2 

•9 

857786  2 

•02 

983067 

4-2, 

0,6933 

7 

54 

69340 

72055 

840985 

2 

\% 

857665  '2 

-o3 

983320 

4-2, 

016680 

6 

55 

69361 

72035 

841116 

3 

857543  2 

.o3 

983573 

4-21 

016427 

5 

56 

69382 

720,5 

841247 

2 

,8 

857422  2 

.o3 

983826 

4-21 

016174 

4 

57 

69403 

7,995 

841378 

2 

,8 

857300  2 

-o3 

984079 

4-21 

015921 

8 

58 

69424 

71974 

841509 

2 

,8 

857178  2 

-o3 

984331 

4-21 

0,5669 

2 

59 

69445 

7,954 

841640 

2 

,8 

857056  '2 

-o3 

984584 

4-21 

oi54,6 

1 

60 

69466 

71934 

841771 

2 

18 

856934  2 

^3 

984837 

4-21 

0,5,63 

0 

N.  COS. 

N.sine. 

L.  COS. 

"d! 

\" 

L.  sine.  | 

L.  cot. 

D.  1" 

L.  tang. 

' 

46° 

1 

74 


TRIGOXOMETRICAL   FL'XCTIOXS. — 44' 


NiT.  Functions. 

Logarithmic  Fcnctioxs  +  10.              1 

0 

X.sine.'N.  cos. 

L.  sine. 

D.  1" 

L.  COS. 

1 

L.  tang. 

Dl." 

L.  cot. 

69466 

71934 

9-841771 

2.18 

9-856034  '2 
856812  '2 

03; 

9.984837 

4.21 

io.oi5i63 

60 

1 

69487 

71914 

841902 

2-l8 

o3 

985090 

4-21 

014910 

59 

2 

69508 

71894 

842033 

2.18 

856690  '.2 

04 

985343 

4-21 

014657 

58 

S 

69329  71873 

842163 

2-17 

856568  2 

04 

985596 

4-21 

014404 

57 

4 

69549  71853 

842294 

2.17 

856446  2 

04 

985848 

4-21 

0i4i52 

56 

5 

69570  71833 

842424 

2-17 

856323  2 

04 

986101 

4-21 

013899 

55 

6 

69591 I71813 

842555 

2-17 

856201  2 

04 

986354 

4-21 

013646 

54 

7 

69612  71792 

842685 

2-17 

856078  2 

04 

986607 

4-21 

013393 

53 

8 

69633171772 

842815 

2-17 

855956  2 
855833  2 

04 

986860 

4-21 

oi3i4o 

52 

9 

69654 '71752 

842946 

2-  17 

04 

9871 12 

4-21 

012888 

51 

10 
11 

69675  71732 

843076 

2-17 

855711  2 

o5, 

987365 

4-21 

01 2635 

50 

69696  7171 I 

9.843206 

2-16 

9.855558  2 

o5 

9.987618 

4-21 

10012382 

49 

12 

69717,71691 

843336 

2-l6 

855465  2 

o5 

987871 
988123 

4-21 

012129 

43 

13 

69737 '7 167 1 

843466 

2-16 

855342  2 

o5 

4-21 

01 1877 

47 

U 

69753,71650 

843595 

2-16 

855219  2 

o5 

988376 

4-21 

011624 

46 

15 

69779*71630 

843725 

2-16 

855096  '2 

o5 

988629 

4-21 

011371 

45 

16 

69800:71610 

843855 

2.16 

854973  2 

o5 

988882 

4-21 

OI1I18 

44 

17 

69821  71390 

843984 

2-16 

854S50  2 

o5 

9S9134 

4-21 

010866 

43 

18 

69842! 71569 

8441 U 

215 

854727  2 

06 

989387 

4-21 

oio6i3 

42 

19 

69862  1  71549 

844243 

2-l5 

834603  I2 

06 

989640 

4-21 

oio36o 

41 

20 

69883 1 71529 

844372 

2.l5 

&544S0  2 

06 

989893 

4-21 

010107 

40 

~2F 

69904  7i5oS 

9-844502 

2-15 

9.854356  2 

06 

Q. 990145 

4-21 

10-009855 

39 

22 

69925 '71488 

844631 

2-l5 

854233  2 

06 

990398 

4-21 

009602 

38 

23 

69946  71468 

844760 

2-l5 

854109  2 

06 

99065 I 

4-21 

009349 

87 

24 

69966 1 71447 

844889 

215 

853986  ,2 

06 

990903 

4-21 

SS 

36 

25 

69987^71427 

845018 

2-15 

853862  12 

06 

991 1 56 

4-21 

35 

26 

70008  71407 

845147 

2-l5 

853738  2 

06 

991409 

4-21 

008591 

34 

27 

70029 '71386 

845276 

2-14 

853614  2 

07 

991662 

4-21 

O08338 

33 

23 

70049  7 1 366 

845405 

2-14 

853490  \7 

07 

991914 

4-21 

608086 

82 

29 

70070; 71345 

845533 

2.14 

853366  12 

07 

992167 

4-21 

007833 

31 

30 

70091 1 7i325 

845662 

2.14 

853242  2 

_o7 

992420 

4-21 

007580 

30 

31 

70112  7i3oj 

9-845790  1  2-14 

9-853ii8  2 

•07 

9.992672 

4-21 

10-007328 

29 

32 

70132:71284 

845919 

2-14 

852994  2 
852869  2 

07 

992925 

4-21 

007075 

28 

83 

70153  71264 

846047 

2-14 

07 

993178 

4-2! 

006822 

27 

34 

70174  71243 

846175 

2-14 

852745  \2 

07 

993430 

4-21 

006570 

26 

35 

70195171223 

846304 

2-14 

852620  2 

07 

993683 

4-21 

oo63i7 

25 

36 

70215  71203 

846432 

2l3 

852496  \2 

08 

993936 

4-2! 

006064 

24 

37 

70236  71 182 

846560 

2.l3 

852371  2 

08 

994189  1  4-21 

oo58ii 

23 

83 

1 70237! 71 162 

846688 

213 

852247  2 

08 

994441  j  4-21 

005559 

22 

89 

70277:71141 

846816  j  2-l3 

852122  '2 

08 

994694   4-21 

oo53o6 

21 

40 

70298  71121 

846944   2-l3 

851997  2 

08 

994947  1  4-21 

oo5o53  1  20  1 

41 

70319  71100 

9-847071  i  2-l3 

9.851872  \2 

"^ 

9.995199  1  4-21 
993452  4-21 

10-004801 

19 

42 

70339  71080 

847199 

2.l3 

851747  12 

08 

004548 

18 

43 

7o36o: 71039 

847327 

2.l3 

85i622  '2 

.08 

995705  '4-21 

004295 

17 

44 

7o38i 1 71039 

847454 

212 

85i497  2 

09 

995937  :  4-21 

004043 

16 

45 

70401  71019 

847582 

2-12 

85i372  2 

09 

996210  ;  4-21 

003790 

15 

46 

70422 ! 70998 

ESS 

2-12 

85 1246  2 

09 

996463 

4-21 

003537 

14 

47 

70443  i  70978 

212 

85ii2i  2 

09 

996715 

4-21 

003285 

13 

48 

70463  i  70957 

847964 
84?ogi 

2-12 

850996  2 

09 

996968 

4-21 

oo3o32 

12 

49 

70484 ' 70937 

2-12 

350870  2 

09 

997221 

4-2! 

002779 

11 

50 

7o5o5  70916 

848218 

212 

850745  2 

oq 

997473 

4-21 

002527 

10 

51 

70523  70896 

9.848345 

212 

9-85o6i9  2 

09 

9-997726  1  4-21 

10-002274  !  9  1 

52 

70546  70875 

848472 

211 

850493  2 

10 

4-21 

002021 

8 

53 

70567  70855 

848599 

2.|I 

830368  2 

10 

998231 

4-21 

001769 

7 

-54 

70587 . 70834 

848726 

211 

85o242  2 

10 

998484 

4-21 

ooi5i6 

6 

55 

70608; 70813 

848852 

2-11 

85oii6  2 

10 

998737 

4-21 

001263 

6 

56 

70628 ,  70793 

848979 

211 

849990  2 

10 

998989 

4-21 

OOIOII 

4 

57 

70649 

70772 

849106 

2-11 

849864  2 

10 

999242 

4-21 

000738 

3 

58 

70670 

70752 

849232 

211 

849738  2 

.10 

999495 

4-21 

000305 

2 

59 

70690 

70731 

849359 
849483 

2-II 

8496 IT  2 

.10 

999747 

4-21 

0002 53 

1 

60 

70711170711 

2-II 

849485  2-10 

lO-OOOOOO  i  4'2I 

10'«00000 

0 

1 

N,  COS.  N.  sine. 

L.  COS.  '  D.  1" 

L.  sine.  ' 

L.  cot.  ,  D.  1" 

L.  tang. 

45° 

1 

TABLE    III. 


PEEOISE    OALCULATIO^f 


FUNCTIONS  NEAR  THEIR  LDIITS. 


76 


SIXES    OF    SMALL    ANGLES. 


log 

.  sin. 

x"=  4.685575  +  log. 

X  -  diff. 

rOR  THE  SIXES 

OF  S3IALL  AXGLES. 

Anarles. 

Second?. 

Diff. 

Angles. 

Seconds 

Diff. 

1 
Angles. 

1 

Seconds. 

Diff. 

o" 

0 

1°  29'  5o" 

5390 

2^  7'3o'' 

765o 

,?-5o'' 

540 

0 

3o'  5o" 

545o    ^r  1 

8'  10" 

7690 

100 

900 

2 
3 

3i'4o" 

55oo 

5; 
53 
54 
55 

8'45" 

7725 

lOl 

102 
io3 

20'  20" 

23' 5o" 

1220 
i43o 

32'  3o" 
33' 3o" 

555o 
56io 

9'  20" 
10' 

7760 
7800 

V' 

1620 

4 
5 

34'  20" 

566o 

10' 40" 

7840 

104 
io5 

29',  5°!! 

1790 

6 

35'  10" 

57.0 

56 

II' i5" 

7875 

106 

33'  3o" 

1900 

36' 

5760 
58io 

ll 

11'  5o" 

7910 

35' 

2100 

7 
8 

36'  5o" 

12' 3o" 

7950 

a 

37' 2C" 

2240 

37' 40' 

5860 

i3'  5" 

7985 

39'  3o ' 

2370 

9 
10 

38'3o" 

5910 

icJ'40' 

8020 

log 
no 

41'  3o" 

2490 

II 
12 

i3 

39' 3o" 

5970 

61 

14' 20" 

8060 

III 
112 

ii3 

114 

ii5 

43'  20" 
45'  10" 

2600 
2710 

40'  20" 
41'  10" 

6020 
6070 

62 

63 

•  64 

65 

i5' 
i5'35" 

8100 
81 35 

47' 

2820 

41' 5o" 

6110 

16'  10" 

8170 

48'  40" 

2920 

14 
i5 

42' 40" 

6160 

16-45" 

8203 

5o' 20" 

3020 

16 

43' 3o" 

6210 

66 

tl 

69 
70 

17' 20" 

8240 

116 

52' 

3l20 

44'  10" 

6200 

17' 55" 

8275 

53'  3o' 

3210 

\l 

45' 

63oo 

i8'3o" 

83io 

VA 

55' 

33oo 

45'  5o" 

63  5o 

19'  5" 

8345 

56' 3o' 

3390 

19 
20 

46' 3o" 

6390 

19' 40" 

838o 

119 

120 

58' 

34S0 

47' 20" 

6440 

71 

74 
75 

20'  1 5" 

8415 

59'  20" 
i<>oo'4o" 

3560 

21 

48' 

6480 

20'  5o" 

845o 

121 
122 

123 
124 
125 

3640 

22 

23 

24 

25 

48'  5o" 

6530 

21' 25" 

8485 

2' 

3720 

49' 3o" 

6570 

22' 

8320 

3'  20- 

38oo 

5o'  20" 

6620 

22' 35" 

8555 

4'  40' 

388o 

26 

5i- 

6660 

76 

23'  10" 

8590 

126 

5'  5o" 

3950 

5i'5o" 

6710 

23' 45" 

8625 

8'  10" 

4020 

27 

28 

29 

3o 

52' 3o" 

6750 

^l 

24  20" 

8660 

4090 

53' 10" 

6790 

24' 55" 

8695 
8730 

9  20" 

■4160 

54' 

6840 

<9 

80 

20' 3o" 

10'  3o" 

4280 

3i 

32 

33 

54'  40" 

6880 

81 

82 
S3 

26' 

8760 

i3i 

l32 

i33 
i34 
i35 

II' 40" 

43oo 

55'  20" 

6920 

26' 35" 

8795 
8825 

12'  5o" 

4370 

56'  10" 

6970 

27'  5" 

14' 
i5' 

4440 
45oo 

34 
35 

56'  5o" 
57'  3o" 

7010 
7o5o 

84 
85 

27' 40" 
28  10" 

8S60 
8890 

16' 10" 

17' 10- 

iTo 

36 

58' ro" 
58'  50* 

?T„ 

86 
ll 

28' 45" 
29'  i5" 

8925 
8955 

1 36 

1 37 

1 38 
139 
140 

18' 10' 

4690 

59'  3o'- 

7170 

29'  5o" 

8990 

19'  20" 
20'  20" 

4760 
4820 

39 
40 

20  00' 10" 
5o" 

7210 
725o 

89 
90 

3o'  20" 
3o'  55" 

9020 
9055 

21' 20" 

4880 

41 

I '40" 

7300 

91 

3i'25" 

oo85 

141 

22'  20" 

4940 

2'  20" 

7340 

o2' 

Q120 

23'  20" 
24'  20" 

25'  10  " 

5ooo 
5o6o 

DIIO 

42 
45 

3' 

3' 35" 
4' 10" 

7380 
7415 
7450 

92 
93 
94 
93 

32' 3o" 
33'  5" 
33' 35" 

9i5o 
9185 
9215 

142 
143 
144 
145 

26' 10  ' 

27  10" 
28'  10" 

5170 

523o 

5290 

46 

47 
48 

49 

4'  5o" 
5'  3o;' 
6'  10" 

7490 
753o 
7570 

96 
ll 

34'  5" 
34'  40" 
35'  10" 

9245 
9280 
9310 

146 

2q 

5340 

6'  5o" 

7610 

35' 40" 

9340 

2q'  5o" 

5390 

7'3o" 

765o 

99 

36' i5" 

9375 

149 

TAXGEXTS    AK"D    C0TA:N^GEXTS   OF   SMALL    AXGLES. 


n 


Jog 

.  tan. 

x"^    4.685575  +  log. 

X  +  diff. 

log 

.  cot. 

x"=  15.314-125  -  log. 

X  -  diff. 

FOR  TANGEi^TS 

AND  COTAi^GENTS  OF 

SMALL  ANGLES. 

Ane^les. 

Seconds. 

Diff. 

Angles. 

Seconds. 

Diff. 

Angles. 

Seconds. 

Diff. 

o" 

0 

1°  3'3o" 

38io 

5o 

5i 

52 

53 

lo3o'io" 

5410 

i  lo" 

43o 

0 

4'  10" 

38D0 

3c' 3o" 

543o 

100 

ii'  lo" 

670 

I 

4'  do" 

3890 

3i' 

5460 

lOI 

14'  10" 

Bdo 

2 

3 

5'3o" 

3930 

3i'3o" 

5490 

102 
io3 

17' 

1020 

6' 

3960 

32' 

5520 

19' 

1440 

4 
5 

6'  40" 

4000 

D4 

55 

32' 20" 

5540 

104 

lOD 

21' 

1260 

6 

7' 20" 

4040 

56 

59 
60 

32' 5o" 

5570 

106 

23' 

i38o 

7'  5o" 

4070 

33' 10" 

5590 

24'  5o" 

1490 

7 
8 

8'3o" 

4110 

33' 40" 

5620 

107 
108 

26' 3o" 

1 590 

9' 

4140 

34'  10" 

565o 

27' 5o" 

i6io 

9 
10 

9'  40" 

4180 

34' 3o" 

5670 

log 
no 

29' 20" 

1760 

T  T 

10' 20" 

4220 

61 
62 

63 

35' 

5700 

3o' 40" 

32' 

1840 
1920 

12 

i3 

10' do" 

ir3o" 

42DO 
4290 

35' 20" 
3d'  5o" 

5720 
5750 

1 1 1 
112 

ii3 

33' 10" 

1990 

12' 

4320 

64 
65 

36'  10" 

5770 

34'  20"  • 

2060 

i4 
i5 

12' 3o" 

43do 

36'  40" 

5800 

114 
Ii5 

35'  3o" 
36'  40" 

2l30 
2200 

16 

i3'io" 
i3' 40" 

4390 
4420 

66 

67 
68 
69 
70 

37' 10" 

37' 3o" 

583o 
585o 

116 

37' 5o" 

2270 

;^ 

14'  10" 

44DO 

3s' 

588o 

]\l 

38' 5o" 

233o 

14' do" 

4490 

38'  20" 

5900 

39' 5o" 

2390 

'9 
20 

i5'  20" 

4D20 

38' do" 

5930 

119 
120 

40' 5o" 

245o 

i5'5o" 

455o 

39' 10" 

5950 

41' do" 

2DIO 

21 

16' 20" 

4d8o 

71 

39'  3o" 

5970 

121 

42'  do" 

2D70 
263o 

22 

23 

24 

25 

17' 

4620 

72 
73 
74 
75 

40' 

6000 

122 

123 

43' do" 

17' 3o" 

465o 

40' 20" 

6020 

44'  40" 

2680 

18' 

4680 

40'  5o" 

6o5o 

124 
125 

45'  40" 

46'  3o" 

2740 
2790 

26 

18' 3o" 
'9' 

4710 
4740 

76 

4i'  10" 
41' 40" 

6070 
6100 

126 

47'  20" 

2840 

27 

28 

19' 3o" 

4770 

77 
78 

42' 

6120 

127 

128 

48'  10" 

2890 

20' 

4800 

42' 3o" 

6i5o 

49' 

2940 

29 

3o 

20'  3o" 

483o 

11 

42'  5o" 

6170 

129 

i3o 

40'  5n" 

2990 

3i 

32 

33 
34 
35 

21' 

4860 

81 
82 
83 
84 
85 

43' 10" 

6190 

i3i 

l32 

i33 
1 34 
i35 

5o'4o" 
5i'3o" 

3040 
3090 

2.'3o" 
22' 

4890 
4920 

43' 40" 
44' 

6220 
6240 

52'  20" 

3 1 40 

22' 3o" 

49D0 

44'  3o" 

6270 

53' 

3i8o 

23' 

49«o 

44'  do" 

6290 

53' 5o" 

323o 

36 

37 
38 
39 

23' 3o" 

5oio 

86 
89 

45'  20" 

6320 

1 36 

III 

54' 40" 

3280 

24' 

5o4o 

45' 40" 

6340 

»  55' 26" 
56' 

3320 
3360 

24'  3o" 

25' 

5070 
5ioo 

46' 
40' 20" 

6360 
6380 

56'  5o" 

3410 

2D' 3o" 

5i3o 

46'  40" 

6400 

139 

40 

90 

140 

57' 3o" 
58'  10" 

34DO 

41 

26' 

5i6o 

47'  10" 

6430 

3490 

26' 3o" 

5190 

91 

47' 3o" 

6450 

141 

58' do" 

3530 

42 
43 
44 
45 

26' 5o" 

52IO 

92 
93 

48' 

6480 

142 
143 

59'  3o" 

3570 

27'  20" 

5240 

48'  20" 

65oo 

1»  0'20" 

3620 

27' Do" 

5270 

94 

9D 

48' 40". 

6520 

144 

■45 

r 

366o 

46 

28' 20" 

53oo 

96 

49' 

6540 

146 

\'  Ao" 

3700 

28' 40" 

5320 

49' 20" 

6560 

2'  10" 

3730 

47 
48 
49 

29'  10" 

535o 

97 
98 

49' 40" 

9580 

',% 

2'  5o" 

3770 

29' 40" 

5380 

5o'  10" 

6610 

3'  3o" 

38 10 

3o'  10" 

5410 

99 

5o'3o" 

663o 

u, 

78 

r 


TAXGEXTS   AND   COTAXGEXTS   OF   SMALL   AXGLES. 


lo^.  tan.  a"=    4.GS5575  +  log.  A+  cliff. 
I02:.  cot.  A"=  15.314425  -  loor.  a-  diff. 


FOR  TAXGEXTS  AXD   COTAXGEXTS    OF   SMALL  AXGLES. 


Anirles. 

^Jfe  do'  3o' 
5o' 5o" 
5i' 10" 
5i'3o" 

D2' 

52' 20" 

5a'  40" 
53' 

53'  20" 
53' do" 
54' 10" 

54' 3o" 
54' 5o" 
55'  10" 
55'  3o" 
55'  5o" 

56'  10" 
56' 3o" 
56'  do" 
57' 10" 
57' 40" 

58' 
58'  20" 

58' 40" 
59' 
59' 20" 

59'  40" 

2°  00' 00" 

20" 

40" 

r 

r  20" 

r4o" 
2' 

2'  20" 
2  40" 

3' 

3'  20" 
3' 40" 
4' 
4'  20" 

4'  40" 
5' 

5'  20" 
5'  40" 
6' 

6'  20" 
6'  40" 

7; 

7  20" 
7  40' 


Seconds.     Diff. 


663o 
66D0 
6670 
669c 
672c 
674c 

6760 
67S0 
6800 
683o 
685o 

6870 
6890 
69 10 
6930 
69D0 

6970 
6990 
7010 
7o3o 
7060 

7080 
7100 
7120 
7140 
7160 

7180 
7200 
7220 
7240 
7260 

7280 
7300 
7320 
7340 
7360 

7380 
^400 
7420 
7440 
7460 

7480 
7000 

7D20 
7340 
7D60 

7580 
7600 
7620 
7640 
7660 


i5o 

IDI 
1D2 

i53 

1D4 
IDD 

1 56 
i57 

IDS 

159 
160 

161 

162 
i63 
164 
165 

166 
167 
168 
169 
170 

171 
172 

173 
174 
175 

176 

177 
178 

179 
180 

18, 
182 
1 83 
184 
i8d 

186 
187 
188 
189 
190 

191 

192 
193 
194 
19D 

196 

197 
198 
199 


Angles. 


7  40' 
8' 

8'i5' 
8'  3o' 

8  5o' 

9  lo' 

9'  3o' 
9'  5o' 
10'  10' 
10'  20' 
10' 40' 

10'  55' 
u'  i5' 
11' 35' 
11'  55' 
12'  i5' 

12' 35' 
12'  55' 
i3'  i5' 
i3'35' 
i3'5o' 

14'  10' 
14'  3o' 
14'  45' 
i5'  5' 
i5'  20' 

i5'4o' 
i5'  55' 
16'  i5' 
16'  3o' 
16'  5o' 

17'  5' 
17'  25' 
17' 40' 
18' 
18' i5' 

18' 35' 
18' 55' 
19'  i5' 
19'  3o' 
19' 45' 

20'  5' 
20  20' 
20'  /iO' 
20'  55" 
21'  i5" 

21' 3o" 
2i'45" 
22'  5" 
22' 20" 
22'  35" 


Seconds. 


7660 
7680 
7695 
7710 
773o 
77D0 

7770 
7790 
7810 
7820 
7840 

7855 
7875 
7895 
791D 
793d 

7955 

7973 
7995 

SoiD 

8o3o 

8o5o 
8070 
8o85 
8io5 
8120 

8i4o 
8i55 
8175 
8190 
8210 

8225 
8245 
8260 
8280 
8295 

83i5 

8335 
8355 
8370 
8385 

8405 
8420 
8440 
8455 
8475 

8490 

&DOD 
8D25 

8540 
8555 


Diff. 


200 

201 
202 
203 
204 
205 

206 
207 
208 
209 
210 

211 
212 
2l3 
214 
2l5 

216 

217 
218 
219 

220 

221 
222 
223 
224 
2  25 

226 
227 
228 
229 
23o 

23l 
232 

233 
234 
235 

236 
237 
238 
239 
240 

241 
242 
243 
244 
245 

246 

247 
248 
249 


Angles, 


2°  22' 35' 
22  55' 
23'  10' 
23' 3c' 
23' 45' 
24' 

24'  20" 
24'  35' 
24'  55' 
25'  10' 

2D'  25" 

25' 45" 
26' 

26'  20" 
26'  35" 
26'  5o" 

27'  10" 
27' 25" 
27' 45" 
28' 
28' i5" 

28' 35" 
28' 5o" 
29' 10" 
29' 25" 
29' 40" 

3o' 

3o'i5" 
3o'  3o" 
3o' do" 
3i'    5" 

3r2o' 
3r35" 
3i'  55' 
32'  10" 

32' 2D" 

32' 40" 
32' 55" 
23' i5" 
33' 3o" 
33' 45" 

34' 

34' i5" 
34' 3o" 
34' 45" 
35' 

35'  20" 
3d' 35" 
35'  5o" 
36'  5" 
36' 20" 


Seconds. 


8555 
8575 
8590 
8610 
8625 
8640 

8660 
8675 
869.N 
8710 
8725 

8745 
8760 
8780 
8795 
8810 

883o 

8845 
8865 
8880 
8895 

8915 
8930 
89D0 
8965 
8980 

9000 
9015 
9o3o 
90D0 
9065 

9080 
909D 
9115 
9i3o 
9145 

9160 
9175 
919D 
921C 
9225 

9240 
9255 
9270 
9285 
9300 

9320 
9335 
9350 
9365 
9380 


TABLE    IV. 


CONTAININQ 


THE  NATURAL  TANGENTS  AND  COTANGENTS 


E^^RY  DEGREE  AND  MINUTE  OF  THE  QUADRANT. 


80 


NATURAL   TAXGEJsTS. 


] 


l  i  0 

1°  1 

r 

3°  1 

4= 

5° 

6° 

7° 

8° 

9° 

n 

0  ,  oooooo 

017455 

034921! 

o524o8 

069927 

087489 

io5io4 

122785 

i4o54i 

158384  60 

291 

746; 

52.2 

2699; 

070219 

7782 

5398 

3980 

0837 

8683  59 

2 

582 

018037 

3281 

55o3 

ODII 

8075 
8368 

5692 

3375 

n34 

8981  58 

3 

873 

5795 

0804 

5987 

3670 
3966 

i43i 

9279  57 

4 

001 164 

619' 

6086 

3575I 

1096 

8661 

6281 

1728 

9577  56 

5 

454 

910 
019201 

6377 

3866 

1389I 

8954 

6575 

4261 

2024 

9876  55 

6 

745 

6668 

41 58 

1681 

9248 

7163 

4557 

2321 

160174  54 

I 

002036 

6960 

445o 

1973 

9541 

4852 

2618 

0472  53 

327 
618 

725, 

4742 

2266 

9832 

7458 

l\% 

2915 

0771  52 

9 

020074 

7542 

5o33 

2558 

090127 

-&% 

3212 

1069  5i 
|368'  5o 

10 

909 

365 

7834 
8125 

5325 

285i 

0421 

5738 

35o8 

003200 

656 

5617 

3i43 

0714 

8340 

7o34 

38o5 

1666  49 
1965  48 

12 

491 

782 

021238' 

8416 

5909 

3435 

1007 

8635 

6329 
6625 

4102 

i3 

8707 

6200 

3728 

i3oo 

8920 
9223 

4399 
4696 

2263  47 

i4 

004072 

529 

8999 

6492 
6784 

4020 

:ii^ 

6920 

2562  46 

i5 

363 

820 

9290 

43i3 

9518 

7216 

4993 

2860  45 

i6 

654'  0221 1 1 

958i 

7076 

46o5 

2180 

9812 

75i2 

5290 
5587 

3i5q  44 
3458  43 

17 

945    402 

9873 

7368 

4898 

2474 

110107 

7807 
8io3 

i8 

005236    6o3 
527    984 

040164 

7660 

5190 
5483 

2767 

0401 

5884 

3756  42 

19 

0456 

Ife 

3o6, 

0695 

8399 

6181 

4o55 

41 

20 

818  023275 

0747 

5775 

3354 

0990 
1284 

8694 

6478 

4354 

40 

21 

006109!    566 

io38 

8535 

6068 

3647 

^ 

X6776 

4652 

ii 

22 

400 

857 
024148 

i33o 

8827 

636i 

3941 

:i?? 

7073 

4951 

23 

t 

1621 

9119 

6653 

4234 

9582 

7370 

525o 

37 

24 

439 

1912 

941 1 

.  6946 

4528 

2168 

9877 

7667 

5540 

36 

23 

007272!    730 

2204 

9703 

7238 

4821 

2463 

130173 

7964 

4848 

35 

26 

563!  025022 

2493 

060287 

7531 

5ii5 

2757 

0469 
0765 

8262 

6147 

34 

27 

854 

3i3 

2787 

7824 
8116 

5408 

3o52 

8559 
8856 

6446  33 

28 

008145 

604 

3^78 

0579 

5702 

3346 

1061 

6745  32 

29 

436 

895 

3370 

0871 

8409 

5995 

3641 

1357 

9154 

7044  3 1 

3o 

727 

026186 

366 1 

ii63 

8702 

6289 
6583 

3936 

i652 

9451 

7343  3o 

3i 

009018.1    477 

3952 

1455 

8994 
9287 

423o 

.  1948 

9748 

7622  29 

32 

^309 

768 

4244 

1747 

6876 

4525 

2244 

1 50046 

7941  28 
8240  27 

33 

600 

027059 

4535 

2039 

9580 

7170 

4820 

2540 

o343 

34 

80. 
010181 

^35o 

4827 

233i 

9873 

7464 

5ii4 

2836 

0641 

8539  26 
8838  25 

35 

641 

5ii8 

2623 

o8oi65 

V^. 

5409 

3x32 

0938 
1236 

36 

472!    933 

541a 

2915 

0458 

5704 

3428 

9137,24 

37 

763'  028224 

5701 

3207 

0751 

8345 

5999 

3725 

1533 

9437  23 

38 

oiio54J    5i5 

^\ 

3499 

1044 

8.638 

6294 

4021 

i83i 

9736  22 

39 

345'    806 

'^oVz 

i336 

8932 

6588 

4317 

2.29 

170035  21 

40 

^^^i  °^'515 

6576 

1629 

9226 

6883 

46i3 

2426 

o334  20 

41 

6867 

4375 

1922 

9519 

7178 

'A2 

2724 

o634  19 
0933  18 

42 

012218:    679 

7159 

4667 

22l5 

9813 

7473 

3022 

43 

509    970 

745o 

4939 

25o8 

J00107 

Vf. 

55o2 

3319 

1233;  17 

44 

800I  030262 

7742 
8o33 

525i 

2B01 

0401 

8o63 

5798 

3617 

i532:  16 

45 

013091 1    553 

5543 

3094 

o6q5 
0989 

8358 

6094 

3915 

i83ii  i5 

46 

382{    844 

8325 

5836 

3386 

8653 

6390 

42i3 

2i3i; i4 

47 

673!  o3ii35 

8617 

6128 

3679 

1282 

8948 

6687 
6983 

45io 

243o  i3 

48 

964J    426 

8908 

6420 

3972 

1576 

9243 

4808 

2730  1 2 

49 

JI4254    717 

9200 

6712 

4265 

1870 

9538 

7279 
7576 

5io6 

3o3o  u 

5o 

545  032009 

1% 

7004 

4558 

2164 

9833 

5404 

3329  lO 

5i 

836i    3oo 

7296 

485 1 

2458 

120128 

7872 

5702 

3629  9 
3920  8 
4228  7 

52 

'%    It 

050075 

7589 

5i44 

2752 

0423 

8169 
8465 

6000 

53 

0366 

7881 
^.73 

5437 

3o46 

0718 

6298 

54 

709'  o33i73 

0658 

5730 

!   3340 

ioi3 

8761 

6596 

4528  6 

55 

016000!   465 

1   0949 

8465 

6023 

1   3634 

^^^ 

9o58 

6894 

4828;  5 

56 

291!    756 

1   1241 

8738 

63i6 

!   3928 

1   i6o4 

9354 

7192 

5l27|4 

57 

582 1  o34o47 

I   i533 

9o5o 

6609 

4222 

1   1899 

965 1 

i£ 

5427' 3 

58 

873    338 
017164    63o 

1824 

9342 

6902 

'   45i6 

'   2194 
2489 

9948 

5427: 2 

^ 

2116 

9635 

7196 

4810 

,  140244 

6027 

I 

89°   88° 

87° 

86° 

85° 

84° 

83° 

82° 

1  81° 

80^ 

Na 

tural  C 

o-tangf 

mts. 

P. 

to 

j:-,  4-CJ 

4-85 

I  4-86 

4-87 

4-88 

1  4-89 

4-91 

4.93 

I4.96 

1  4-98 

NATTJKAL    TANGENTS. 


81 


is 

10° 

11° 

12° 

13° 

14° 

15° 

16° 

17° 

18° 

19° 

"1 

176327 

194380 

212557 

23o868 

249328 

267949 

286745 

3o573i 

324920 

344328'  60 

6627 

4682 

2861 

1175 

9637 

8261 

7060 

6049 

5241 

4653,  59 
4978:  58 

6927 

4984 

3i65 

i48i 

9946 

8573 

7375 

6367 

5563 

7227 

5286 

3469 

1788 

250255 

8885 

& 

6685 

5885 

53o4i  57 

7527 

5588 

3773 

2094 

o564 

9197 

7003 

6207 

563o  56 

7827 
8127 

5890 

4077 

2401 

0873 

9309 

8320 

7322 

6528 

5955 

55 

6 

6192 

4381 

2707 

Ii83 

9821 

8635 

7640 

685o 

6281 

54 

I 

8427 

6494 

4686 

3oi4 

1492 

270133 

8950 

7959 
8277 

7172 

6607!  53 

8727 

6796 

4990 

3321 

1801 

0445 

9266 

7494 

6933  52 

9 

9028 

7099 

5294 

3627 

2111 

0737 

9581 

8396 

7817 
328139 

7259 

5i 

10 

179328 

I 9740 I 

215599 

233934 

252420 

271069 

289896 

308914 

347583 

5o 

II 

9628 

77o3 

5903 

4241 

2729 

i382 

290211 

9233 

8461 

79" 

49 

12 

9928 

8oo5 

6208 

4548 

3039 
3348 

1694 

0327 

9552 

8783 

8237' 48 

i3 

180229 

83o8 

65i2 

4855 

2006 

0842 

9871 
310189 

9106 

8563'  47 

14 

0329 

8610 

6817 

4162 

3658 

2319 

ii58 

9428 

8889' 46 

i5 

0829 

8912 

7121 

5469 

3968 

263 1 

1473 

o5o8 

975i 

92i6|43 

i6 

ii3o 

92i5 

7426 

5776 

4277 

2944 

1789 

0827 

330073 

9542  44 

\l 

i43o 

9517 

7731 

5o83 

4587 

3256 

2103 

1 146 

0396 

9868  43 

I73i 

9820 

8o35 

6390 

4897 

3569 

2420 

1465 

0718 

3501951  42 

19 

2o3i 

200122 

8340 

6697 

5207 
2555i6 

3882 

2736 

1784 

1 041 

0522 

41 

20 

182332 

200425 

218645 

237004 

274194 

293o52 

312104 

33i364 

350848:  40 

21 

2632 

0727 

8950 

73.2 

5826 

4307 

3368 

2423 

1687 

1175  39 

22 

2933 

io3o 

9254 

7619 

6i36 

4820I   3684 

2742 

2010 

i5o2'38 

23 

3234 

1333 

9559 

IVm 

6446 

5i33 

4000 

3062 

2333 

1829' 37 

24 

3534 

1635 

9864 

6756 

5446 

43i6 

338i 

2656 

2156; 36 

25 

3835 

1938 

220169 

8541 

7066 

5759 

4632 

3700 

2679 

24831  35 

26 

4i36 

2241 

0474 

8^8 

7377 

6072 

4948 

4020 

3302 

2810' 34 

27 

4437 

2544 

0779 
1084 

9156 

7687 

6385 

5265 

4340 

3623 

3i37;33 

28 

4737 

2847 

9464 

7997 

6698 

558i 

4659 

3949 

3464'  02 

29 

5o38 

3149 

1 389 

9771 

83o7 

701 1 

5897 

4979 

4272 

379i|3i 

3o 

185339 

203432 

221693 

240079 

2586i8 

277325 

296213 

3i5299 

334393 

354119:  3o 

3i 

5640 

3755 

2000 

o386 

8928 

7638 

653o 

5619 

4919 

4446|  29 

32 

5941 

4o58 

23o5 

0694 

9238 

795 1 

6846 

5939 

5242 

4773 

28 

33 

6242 

436i 

2610 

1002 

9549 

8265 

7163 

6258 

5566 

5ioi 

27 

34 

6543 

4664 

2916 

i3io 

9859 

8578 

7480 

6578 

5890 

5429'  26 

35 

6844 

4967 

3221 

1618 

260170 

8891 

7796 

6899 

62i3 

5756I  25 

36 

7145 

5271 

3526 

1925 

0480 

9205 

8ii3 

7219 

6537 

6084  24 

ll 

7446 

5574 

3832 

2233 

0791 

9519 

843o 

7539 

6861 

6412  23 

8048 

lUl 

4i37 

2341 

1102 

9832 

8747 
9063 

7839 
8179 

7i85 

6740  22 

39 

4443 

2849 

i4i3 

280146 

75og 

7068  21 

357396  20 

40 

188349 

206483 

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5i33 

881 3 

2699 

6741 

35 

26 

2070 

2653 

6756 

980413 

i524 

8876 

2765 
283 1 

6809 
6878 

34 

27 

2573 

3i7i 

4727 

7307 

0983 

1 583 

5194 

8940 

33 

28 

3075 

3689 

5261 

7859 

i554 

1642 

5255 

9003 

2897 

6947 

32 

29 

3578 

4207 

5796 

8412 

2126 

1702 

53,7 

9067 

2963 

7016 

3i 

30 

854081  884725 

9i633i 

948965 

982697 

1-01761 

1.05378 

1.09131 

I- 13029 

1-17085 

3o 

3i 

4584 

5244 

6866 

9318 

3269 

1820 

5439 

9195 

3096 

7154 

11 

32 

5087 

5763 

7402 

950071 

3842 

1879 

55oi 

9238 

3x62 

7223 

33 

5591 

6282 

7938 

0624 

4414 

2037 

5562 

9322 

3228 

7292 

27 

34 

6095 

6802 

8474 

1178 

4987 

5624 

9386 

3295 

7361 

26 

35 

6599 

7321 

9010 

1733 

5560 
fei34 

5685 

945o 

336i 

743oj  23 

36 

7104 

7842 

8362 

9547 

2287 

2117 

5747 
5^09 

9514 

3428 

75oo  24 

37 

7608 

920084 

2842 

6708 

2176 

9578 

3494 

756Q|  23 

38 

81 13   8882 

0621 

3397 

7282 

2236 

5870 

9642 

356i 

7638 

22 

39 

8619!   9403 

ii59 

3953 

7857 
988432 

2295 

5932 

9706 

3627 

7708 

21 

40 

859124;  889924 

921697 

954308 

1-02355 

1-05994 

1-09770 

1.13694 

1-17777 

20 

41 

963o 

890446 

2235 

5o64 

9007 

24i4 

6036 

9834 

3761 

7846 

\t 

42 

8601 36 

0967 

2773 

5621 

9582 

lili 

6117 

9899 

3828 

7916 

43 

0642 

1489 

3312 

6177 
6734 

990158 

6179 

9963 

3894 

7986 

17 

44 

1148 

2012 

385i 

0735 

2593 

6241 

1-10027 

3961 

8o55 

16 

45 

1655 

2534 

4390 
49^0 

7292 

i3ii 

2653 

63o3 

0091 
01 56 

4028 

8i25 

i5 

46 

2162 

3o57 

7849 

1888 

2713 

6365 

4095 

8194 

14 

47 

2669 

358o 

5470 

8407 

2465 

im 

6427 

0220 

4162 

8264 

i3 

48 

Ull 

4io3 

6010 

8966 
9524 

3o43 

6489 

0285 

4229 

8334 

12 

49 

4627 

655i 

3621 

2892 

6551 

0349 

4296 

8404'  11 1 

5o 

864193 

895i5i 

8174 

960083 

994199 

1-02932 

1.06613 

1.10414 

I- I 4363 

1-18474,  10 1 

5i 

4701 

5675 

0642 

4778 

3012 

6676 

0478 

443o 

8544 

t 

52 

5209 

5718 

6199 

1202 

5357 

3072 

6738 

o543 

4498 
4565 

8614 

53 

6724 

8715 

1761 

5936 

3i32 

6800 

0607 

8684 

I 

54 

6227 

7249 

9257 

2322 

65i5 

3192 

6862 

0672 

4632 

8754 

55 

6736 

7774 

9800 

2882 

7095 

3232 

6925 

0737 

4699 

8824 

5 

56 

7246 

8299;  93o342 

3443 

7676 

33l2 

6987 

0802 

tltl 

8894 

4 

u 

2756 
8266 

8823 

o885 

4004 

8256 

3372 
3433 

7049 

0867 

8964 

3 

9351 

1428 

4565 

8837 
9418 

7112 

0931 

4902 

9035 

2 

59 

8776 

9877 

1971 

5127 

3493 

7174 

0996 

4969 

9105 

1 

49° 

48° 

47' 

46° 

45° 

44° 

43° 

42° 

41° 

40° 

Natural  C 

0- tangents. 

ro?^«-38 

8-64   8-92 

9-21   9-53 

O-OQ 

1. 02 

1.06 

I>10 

,.,5 

] 


NATURAL    TANGENTS. 


85 


ifi. 

50° 

51° 

52° 

53° 

54° 

55° 

56° 

57° 

58° 

59° 

n 

0 

1-19175 

1.23490 

1-27994 

1-32704 

1-37638 

i.428i5'i-48256 

1-53986 

i-6oo33 

1-66428  60 

I 

9246 

3563 

8071 

2785 

7722 

2903 

8349 

4o85 

0137 

6538  59 
6647;  58 

2 

9316 

3637 

8148 

2865 

7807 

2992 

8442 

4i83 

0241 

3 

9387 

3710 

8225 

2946 

7891 

3o8o 

8536 

4281 

0345 

6757  57 

4 

nil 

3784 
3858 

83o2 

3026 

^ 

3169 
3258 

8629 

4379 
4478 

0449 
o553 

6867  56 
6978  55 
7088  54 

5 

lilt 

3io7 

8722 
8816 

6 

9599 

3931 

3187 

8145 

3347 

4576 

0657 

I 

9669 

4oo5 

8533 

3268 

8229 

3436 

8909 
90o3 

4675 

0761 

7198,53 

9740 

iVi 

8610 

3349 

83i4 

3525 

4774 

o865 

7309  52 

9 

9811 

8687 

3430 

i.384?4 

36i4 

9097 

4873 

0970 

7419  5i 

10 

I. 19882 

1-24227 

1.28764 

I-335I1 

1-437031.49190 

1.54972 

1.61074 

1-67530  5o 

9953 

43oi 

8842 

3592 

8658 

3792 

9284 

507. 

'AM 

7641149 
7752,  48 

12 

1-20024 

4375 

8919 

3673 

8653 

388i 

9378 

§170 

i3 

0095 

444Q 
4523 

8997 

3754 

8738 

3970 

9472 

5269 
5368 

1388 

7863'  47 

14 

0166 

9074 
9i52 

3835 

8824 

4060 

9566 

1493 

79741  46 
8o85  45 

i5 

0237 

4597 

3916 

8909 

4i49 

9661 

5467 

1598 

i6 

o3o8 

4672 

9229 

3998 

8994 

5239 

9755 
9849 

5567 

I2o8 

8196  44 

\l 

0379 

4746 

9307 

4079 

mt 

432Q 

4418 

5666 

83o8!  43 

045 1 

4820 

9385 

4160 

9944 

5766 
5866 

1914 

8419  42 

'9 

0522 

4895 

9463 

4242 

9230 

45o8 

i.5oo38 

2019 

8531I41 

20 

1-20593 

1-24969 

1-29541 

1-34323 

1-39336 

1-44598 
4688 

i.5oi33 

1.55966 

1.62123 

1-68643;  40 

21 

o665 

5044 

9618 

44o5 

9421 

02  2B 

6o65 

223o 

8754  39 
8866;  38 

22 

0736 
0808 

5ii8 

9696 

4487 

9507 

4778 
486S 

0322 

6i65 

2336 

23 

5193 

9775 

4568 

9593 

0417 

6265 

2442 

8979'  37 

24 

0879 

5268 

9853 

4650 

9679 

4958 

05l2 

6366 

2548 

9091136 

25 

0951 

5343 

9931 

4732 

9764 

5o49 

0607 

6466 

2654 

9203  35 

26 

1023 

5417 

1-30009 

4814 

985o 

5i39 

0702 

6566 

2760 

9316'  34 

27 

1094 

5492 

0087 

4896 

9936 

5229 

0797 

6667 

2866 

9428:  33 

28 

1166 

5567 

0166 

.  497« 

1.40022 

5320 

0893 
0988 

t&l 

2972 

9541  32 

29 

1238 

5642 

0244 

5o6o 

0109 

5410 

3079 

9653' 3 1 

3o 

I.2l3lO 

1-25717 

1-30323 

I-35I42 

1-40195 

I -45501 

1.51084 

1-56969 

i.63i8D 

i.69766|3o 

3i 

i382 

5792 

0401 

5224 

0281 

5592 

1170 
1275 

7069 

3292 

9879  20 
9992  28 

32 

1454 

5867 

0480 

5307 

0367 

5682 

7170 

3398 

33 

i526 

5943 

0558 

5389 

0454 

lul 

1370 

7271 

35o5 

I. 70 106  27 

34 

1598 

6018 

0637 

5472 

o54o 

1466 

7372 

3612 

0219  26 

35 

1670 

6093 

0716 

5554 

0627 

5955 

i562 

7474 

3719 
3826 

o332  25 

36 

1742 

6169 

0695 

5637 

0714 

6046 

1658 

7575 

0446  24 

11 

1814 

6244 

0873 

5719 

0800 

6i37 

i85o 

7676 

3934 

o56o;  23 

1886 

63i9 
639! 

0952 

5802 

0887 

6229 

7778 
7879 

4041 

0673  22 
0787  21 

.39 

1959 

io3i 

K   5885 

0074 

6320 

1946 

4148 

40 

I-2203l 

I -2647 1 

I-3III0 

1.35968 

1-41061 

1.46411 

1.52043 

'•% 

1-64256 

1.70901  20 

41 

2104 

6546 

1190 

6o5i 

1 148 

65o3 

2139 

4363 

ioi5!  19 

42 

2176 

6622 

1269 
1348 

6i34 

1235 

6595 

223d 

8184 

4471 

II29'l8 

43 

2249 

6698 

6217 

l322 

6686 

2332 

3286 

4579 

1244!  17 

44 

2321 

6774 

1427 

63oo 

1409 

6778 

2429 

8388 

4687 

i358l  16 

45 

2394 

6849 
6925 

1 507 

6383 

1497 

6870 

252D 

8490 

4795 

1473  i5 

46 

2467 

1 586 

6466 

1 584 

6962 

2622 

8593 

4903 

i588l  14 

tl 

2539 

7001 

1666 

6549 

1672 

7053 

lU 

8695 

Soil 

1702J  i3 

2612 

7077 

1745 

6633 

1759 

7146 

8797 

5l20 

i8i7j  12 

49 

2685 

7153 

1825 

6716 

1847 

7238 

2913 

8900 

5228 

1932'  II 

5o 

1-22758 

1-27230 

1-31904 

1-36800 

1-41934 

1-47330 

I.530IO 

1-59002 

1-65337 
5445 

1.72047I  10 

5i 

283 1 

7306 

1984 

6883 

2022 

7422 

3107 
3205 

9105 

2163 

I 

52 

.2904 

7382 

2064 

6967 

2110 

7514 

9208 

5554 

2278 

-63 

2977 

7458 

J144 

7o5o 

2198 
2286 

7607 

3302 

9311 

5663 

2393 

7 

54 

3o5o 

7535 

2224 

7134 

7699 

3400 

9414 

5772 

262: 

6 

55 

3i23 

7611 

23o4 

7218 

2374 

]lll 

3497 

95,7 

5881 

5 

56 

3196 

7688 

2384 

7302 

2462 

3595 

9620 

5990 

2741 
2857 
2973 

4 

57 

3270 

7764 
7841 

2464 

7386 

255o 

7977 
8070 

3693 

9723 
9826 

6099 

3 

58 

3343 

2544 

7470 

2638 

3791 

6209 

2 

59 

3416 

7917 

2624 

7554 

2726 

8i63 

3888 

9930 

63i8 

3089 

1 



39* 

38' 

37° 

30° 

35- 

34° 

33° 

32° 

31° 

30° 

a 

Natural  Co-tangeuts. 

P. 

to 

v;  1-20 

1-25 

i.3i 

1-37 

1-44 

i.5i 

1.59 

1-68 

1.78 

1.88 

86 


NATURAL    TANGENTS. 


60° 

6r 

62° 

63° 

64° 

65°   66° 

67° 

68° 

69° 

~\ 

1.73205 

i.8o4o5 

1.88073 

1-96261  2.o5o3o 

2.i445i  2-24604 2-35535.2-47509  2-6o5o9  6o| 

3321 

o529 

8205 

6402 

5i82 

4614   4780 

5776 

7716 

0736  59 
0963,  5§ 

3438 

o653 

8337 

6544 

5333 

4777 

4956 

5967 

3924 
8i32 

3555 

0777 

8469 

6685 

5485 

4940 

5i32 

61 58 

1190  57 

3671 

0901 

8602 

6827 

5637 

5io4 

5309 

6349 

8340 

1418  56 

5 

8788 

To25 

8734 

6969 

5790 

5268 

5486 

6541 

8549 
8758 

1646  55 

6 

3905 

ii5o 

8867 

7111 

5942 

5432 

5663 

6733 

1874  54 

I 

4022 

1274 

9000 

7253 

6094 

5596 

5840 

6925 

8967 

2io3  53 

4140 

,399 

9133 

7395   6247 
7538   6400 

5760 

6018   7118 

l^sl 

2332'  52 

9 

4257 

1 524 

9266 

5925 

6196   73ii 

256i;5i 

lO 

1-74375 

1-81649 

1.89400 

1-97681  2-06553 

2 .  16090  2 •  263742 - 37504 
6255   65521   7697 

2.49597  2-62791!  5o| 

II 

4492 

1774 

9533 

7823 

6706 

9807 

3o2i!  49 
3252  48 

12 

4610 

1899 

9667 

7966 
8110 

6860 

6420   6730 

IX 

2-50018 

i3 

4728 

2025 

9801 

7014 

6585   6909 
6751   7088 

0229 

3483  47 

14 

4846 

2i5o 

9935 

8253 

7167 

t?,t 

0440 

37i4i  46 

i5 

4964 

2276 

1.90069 

0203 

8396 

7321 

6917   7267 

o652 

3945;  45 

i6 

5082 

2402 

8540 

7476 

7083   7447 

8668 

0864 

4177  44 

\l 

5200 

2528 

0337 

8684 

763o 

7249   7626 

8863 

1076J   4410^  43 
1289   4642  42 

53i9 

2654 

0472 

8828 

7785 

7416   7S06 

9o58 

19 

5437 

2780 

0607 

8972 

7939 

7582   7987 

9253 

i5o2   4875,41 

20 

1.75556 

1-82906 

I. 90741 

I -99116 

2 • 08094 

2-17749  2-28167  2-39449|2.5i7i5  2-65109  40  | 

21 

5675 

3o33 

0876 

■  9261 

825o 

7916;   8348i   9645 

1929 

5342  39 
5576  38 

22 

5794 

3i59 

I0I2 

9406 

8405 

8084 

8528|   9841 

2142 

23 

59T3 

3286 

1 147 

9550 

856o 

825i 

87102-40038 

2357 

58111  37 

24 

6o32 

34i3 

1282 

9695 

8716 

8419 

8891 

0235 

2571 

6046,  36 

25 

6i5i 

3540 

1418 

9S41 

8872 

8587 

9073 

0432 

2786 

6281;  35 

26 

6271 

3667 

1554 

9986 

9028 

8755 

9254 

0629 

3ooi 

65i6  34 

U 

6390 

3794 

1690 

2-OOl3l 

9184 

8923 

9437 

0827 

3217 

6752  33 

65x0 

3922 

1826 

0277 

9341 

9092 

9619 

1025 

3432 

6989'  32 

29 

6630 

4049 

1962 

0423 

9498 

9261 

9801 

1223 

3648 

7225,3. 

3o 

1.76749 

1-84177 

'■Kt 

2-00569 

2-09654 

2-19430  2-29984|2-4i42i 

2 -538652 -67462!  3o| 

3i 

6869 

43o5 

0715 

0862 

9811 

95992-30167 

1620 

4082 

7700I  29 
79371  28 
8175;  27 

32 

6990 

4433 

2371 

9969 

9769 
9938 

o35i 

1819 

4299 

33 

7110 

4561 

25o8 

1008  2.10126 

0534 

2019 
2218 

45i6 

34 

7230 

4689 
4818 

2645 

1155 

0284 

2-20108 

0718 

4734 

8414I  26 

35 

7351 

2782 

I302 

0442 

0278 

0902 

2418 

4952 

8653!  25 

36 

7471 

4946 

2920 

1449 

0600 

0449 

1086 

2618 

5170 
5389 
56o8 

8802;  24 
9i3ii  23 

11 

7592 

5075 

3o57 

1596 

0708 

0619 

1271 

2819 

7713 
7834 

5204 

3io5 
3332 

1743 

0916 

0790 
0961 

1456 

3019 

937I]  22 

39 

5333 

1801 
2-02039 

1075 

1641 

3220 

5827 

9612  21 

40 

1.77955 

1.85462 

1-93470 

2-11233 

2-21132  2.3l826'2-43422 

2- 56o46;2- 698531  20 1 

41 

8077 
8198 

5591 

36o8 

2187 

1392 

l3o4    2012 

3623 

6266,2.70094'  iqI 

42 

5720 

3746 
3885 

2335 

1552 

1475 

nil 

3825 

6487 

o335  18 

43 

83i9 

585o 

2483 

1871 
2o3o 

1647 

4027 

6707 

05771  17 

44 

8441 

5979 

4023 

263 1 

1819 

2570 

423o 

6928 

0819!  16 

45 

8563 

6109 

4162 

2780 

1992 

2756   4433 

7.5o 

1062  i5 

46 

8685 

6239 

43oi 

2929 

2190 

2164 

2943   4636 

7371 

i3o5  14 

% 

8807 

6369 

4440 

3078 

23  DO 

2337 

3i3o   4839 
3317   5o43 

7593 

1548  i3 

8929 

6499 

4570 

4718 

1-94858 

3227 

25ll 

25l0 

7815 
8o38 

1792:  12 
2o36|  11 

49 

905 1 

663o 

3376 

2671 

2683 

35o5   5246 

5o 

1.79174 

1.86760 

2-03526 

2-12832 

2-22857 

2-336932-45451 

2.58261 

2.72281  10 

5i 

9296 

6891 

iV3] 

3675 

2993 

3o3o 

388 1   ~ 

5655 

8484 

2526 

I 

52 

9419 

7021 

3825 

3Id4 

3204 

T.^ 

586o 

8708 

2771 

53 

9542 

7i52 

5277 

3975 

33i6 

3378 

6o65 

8932 

3017 

7 

54 

9665 

7283 

54.7 

4125 

3477 

3553 

4447 

6270 

91 56 

32631  6 

55 

9788 

74i5 

I 

4276 

3639 

3727 

4636 

6476 

9381 

3509  5 

3756i  4 

56 

9911 

7546 

4426 

38oi 

3902 

4825 

6682 

9606 

5? 

I -80034 

7677 

58?8 

4577 

3963 

4077 

5oi5 

6888 

9831 

4004!  3 

58 

01 58 

7809 

5979 

4728 

4125 

4252 

52o5 

70952-60057 

4i5i 

2 

59 

0281 

7941 

6120 

4879 

4288 

4423 

5395 

73oa 

0283 

4499 

_i 

29° 

2S° 

27° 

26° 

25° 

24° 

23° 

22° 

21° 

20° 

3 

Natural  C 

o-tangeuts. 

P. 
to] 

^;>.oo 

2-13 

2.27 

1-44 

i.62 

a-82 

3-05 

331 

3-6i 

3.95 

NATUEAL    TANGENTS. 


87 


s 

70° 

71°   72° 

73° 

74° 

75°   76° 

77° 

78°   70°  1 

0 

2-74748  2.9042i;3-07768 

3-27085 

3-48741  3-732054-01078 

4-33148 

4-70463  5-14455  60 

I 

4997 

0696   8073 

7426 

9125 

3640   1576 

3723 

1137   5236  5q 

2 

5246 

0971 

8379 

7767 

9509 

4075 

2074 

43oo 

i8i3 

6o58  58 

3 

5496 

1246 

8685 

8109 

9894 

45i2 

2574 

4879 

2490 

6863!  57 

4 

5746 

i523 

8991 

8452 

3-50279 

4930 

3076 

5459 

3170 

7671  56 

5 

5996 

1799 

9298 

8795 

0666 

5388 

3578 

6040 

3S5i 

8480  55 

6 

6247 
6498 
67D0 

2076 

9606 

9139 

io53 

5828 

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.... 

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.... 

1 


'9 


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OOLTON'S   NEW  GEOGEAPHIES. 

Hie  tchole  subject  in  Two  Jioohs. 

These  hooka  are  the  most  simple,  the  most  practical,  and  beat 
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Elegantly  Illustrated. 

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raphy that  is  practical  and  available  during  the  t»hort  time  we  have  used  this 
work,  than  in  all  my  life  before,  including  ten  years  teaching  by  Mitcheirs 
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So  well  satisfied  have  I  been  with  these  Geographies  that  I  adopted  them, 
and  have  procured  their  introduction  into  most  of  the  schools  in  this  county. 
James  W.  Thompson,  A.M.,  Prin.  of  CentreHlle  Academy,  Maryland. 

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